Pythagorean triple

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Animation demonstrating the smallest Pythagorean triple, 3² + 4² = 5².

A Pythagorean triple consists of three positive integers a, b, and c, such that a² + b² = c². Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are coprime (that is, they have no common divisor larger than 1). For example, (3, 4, 5) is a primitive Pythagorean triple whereas (6, 8, 10) is not. A triangle whose sides form a Pythagorean triple is called a Pythagorean triangle, and is necessarily a right triangle.

The name is derived from the Pythagorean theorem, stating that every right triangle has side lengths satisfying the formula $a^2+b^2=c^2$; thus, Pythagorean triples describe the three integer side lengths of a right triangle. However, right triangles with non-integer sides do not form Pythagorean triples. For instance, the triangle with sides $a=b=1$ and ${\displaystyle c=\sqrt{2}}$ is a right triangle, but ${\displaystyle (1,1,\sqrt{2})}$ is not a Pythagorean triple because $\sqrt{2}$ is not an integer. Moreover, $1$ and $\sqrt{2}$ do not have an integer common multiple because $\sqrt{2}$ is irrational.

Pythagorean triples have been known since ancient times. The oldest known record comes from Plimpton 322, a Babylonian clay tablet from about 1800 BC, written in a sexagesimal number system. It was discovered by Edgar James Banks shortly after 1900, and sold to George Arthur Plimpton in 1922, for $10.

When searching for integer solutions, the equation a² + b² = c² is a Diophantine equation. Thus Pythagorean triples are among the oldest known solutions of a nonlinear Diophantine equation.

Examples

Scatter plot of the legs (a, b) of the first Pythagorean triples with a and b less than 6000. Negative values are included to illustrate the parabolic patterns. The "rays" are a result of the fact that if (a, b, c) is a Pythagorean triple, then so is (2a, 2b, 2c), (3a, 3b, 3c) and, more generally, (ka, kb, kc) for any positive integer k.

There are 16 primitive Pythagorean triples of numbers up to 100:

(3, 4, 5)	(5, 12, 13)	(8, 15, 17)	(7, 24, 25)
(20, 21, 29)	(12, 35, 37)	(9, 40, 41)	(28, 45, 53)
(11, 60, 61)	(16, 63, 65)	(33, 56, 65)	(48, 55, 73)
(13, 84, 85)	(36, 77, 85)	(39, 80, 89)	(65, 72, 97)

Other small Pythagorean triples such as (6, 8, 10) are not listed because they are not primitive; for instance (6, 8, 10) is a multiple of (3, 4, 5).

Each of these points (with their multiples) forms a radiating line in the scatter plot above.

Additionally, these are the remaining primitive Pythagorean triples of numbers up to 300:

(20, 99, 101)	(60, 91, 109)	(15, 112, 113)	(44, 117, 125)
(88, 105, 137)	(17, 144, 145)	(24, 143, 145)	(51, 140, 149)
(85, 132, 157)	(119, 120, 169)	(52, 165, 173)	(19, 180, 181)
(57, 176, 185)	(104, 153, 185)	(95, 168, 193)	(28, 195, 197)
(84, 187, 205)	(133, 156, 205)	(21, 220, 221)	(140, 171, 221)
(60, 221, 229)	(105, 208, 233)	(120, 209, 241)	(32, 255, 257)
(23, 264, 265)	(96, 247, 265)	(69, 260, 269)	(115, 252, 277)
(160, 231, 281)	(161, 240, 289)	(68, 285, 293)

Generating a triple

Main article: Formulas for generating Pythagorean triples

The primitive Pythagorean triples. The odd leg a is plotted on the horizontal axis, the even leg b on the vertical. The curvilinear grid is composed of curves of constant m − n and of constant m + n in Euclid's formula.

A plot of triples generated by Euclid's formula maps out part of the z² = x² + y² cone. A constant m or n traces out part of a parabola on the cone.

Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of integers m and n with m > n > 0. The formula states that the integers

$a=m^2-n^2,b=2mn,c=m^2+n^2$

form a Pythagorean triple. The triple generated by Euclid's formula is primitive if and only if m and n are coprime and one of them is even. When both m and n are odd, then a, b, and c will be even, and the triple will not be primitive; however, dividing a, b, and c by 2 will yield a primitive triple when m and n are coprime.

Every primitive triple arises (after the exchange of a and b, if a is even) from a unique pair of coprime numbers m, n, one of which is even. It follows that there are infinitely many primitive Pythagorean triples. This relationship of a, b and c to m and n from Euclid's formula is referenced throughout the rest of this article.

Despite generating all primitive triples, Euclid's formula does not produce all triples—for example, (9, 12, 15) cannot be generated using integer m and n. This can be remedied by inserting an additional parameter k to the formula. The following will generate all Pythagorean triples uniquely:

$a=k\cdot (m^2-n^2),b=k\cdot (2mn),c=k\cdot (m^2+n^2)$

where m, n, and k are positive integers with m > n, and with m and n coprime and not both odd.

That these formulas generate Pythagorean triples can be verified by expanding a² + b² using elementary algebra and verifying that the result equals c². Since every Pythagorean triple can be divided through by some integer k to obtain a primitive triple, every triple can be generated uniquely by using the formula with m and n to generate its primitive counterpart and then multiplying through by k as in the last equation.

Choosing m and n from certain integer sequences gives interesting results. For example, if m and n are consecutive Pell numbers, a and b will differ by 1.

Many formulas for generating triples with particular properties have been developed since the time of Euclid.

Proof of Euclid's formula

That satisfaction of Euclid's formula by a, b, c is sufficient for the triangle to be Pythagorean is apparent from the fact that for positive integers m and n, m > n, the a, b, and c given by the formula are all positive integers, and from the fact that

$a^2+b^2=(m^2-n^2{)}^2+(2mn{)}^2=(m^2+n^2{)}^2=c^2.$

A proof of the necessity that a, b, c be expressed by Euclid's formula for any primitive Pythagorean triple is as follows. All such primitive triples can be written as (a, b, c) where a² + b² = c² and a, b, c are coprime. Thus a, b, c are pairwise coprime (if a prime number divided two of them, it would be forced also to divide the third one). As a and b are coprime, at least one of them is odd, so we may suppose that a is odd, by exchanging, if needed, a and b. This implies that b is even and c is odd (if b were odd, c would be even, and c² would be a multiple of 4, while a² + b² would be congruent to 2 modulo 4, as an odd square is congruent to 1 modulo 4).

From $a^2+b^2=c^2$ we obtain $c^2-a^2=b^2$ and hence $(c-a)(c+a)=b^2$. Then $\frac{(c+a)}{b}=\frac{b}{(c-a)}$. Since $\frac{(c+a)}{b}$ is rational, we set it equal to $\frac{m}{n}$ in lowest terms. Thus ${\displaystyle \frac{(c-a)}{b}=\frac{n}{m}}$, being the reciprocal of $\frac{(c+a)}{b}$. Then solving

$\frac{c}{b}+\frac{a}{b}=\frac{m}{n},\frac{c}{b}-\frac{a}{b}=\frac{n}{m}$

for $\frac{c}{b}$ and $\frac{a}{b}$ gives

$\frac{c}{b}=\frac{1}{2}\left(\frac{m}{n}+\frac{n}{m}\right)=\frac{m^2+n^2}{2mn},\frac{a}{b}=\frac{1}{2}\left(\frac{m}{n}-\frac{n}{m}\right)=\frac{m^2-n^2}{2mn}.$

As $\frac{m}{n}$ is fully reduced, m and n are coprime, and they cannot both be even. If they were both odd, the numerator of ${\displaystyle \frac{m^2-n^2}{2mn}}$ would be a multiple of 4 (because an odd square is congruent to 1 modulo 4), and the denominator 2mn would not be a multiple of 4. Since 4 would be the minimum possible even factor in the numerator and 2 would be the maximum possible even factor in the denominator, this would imply a to be even despite defining it as odd. Thus one of m and n is odd and the other is even, and the numerators of the two fractions with denominator 2mn are odd. Thus these fractions are fully reduced (an odd prime dividing this denominator divides one of m and n but not the other; thus it does not divide m² ± n²). One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula

$a=m^2-n^2,b=2mn,c=m^2+n^2$ with m and n coprime and of opposite parities.

A longer but more commonplace proof is given in Maor (2007) and Sierpiński (2003). Another proof is given in Diophantine equation § Example of Pythagorean triples, as an instance of a general method that applies to every homogeneous Diophantine equation of degree two.

Interpretation of parameters in Euclid's formula

Suppose the sides of a Pythagorean triangle have lengths m² − n², 2mn, and m² + n², and suppose the angle between the leg of length m² − n² and the hypotenuse of length m² + n² is denoted as β. Then ${\displaystyle \tan \frac{\beta }{2}=\frac{n}{m}}$ and the full-angle trigonometric values are ${\displaystyle \sin \beta =\frac{2mn}{m^2+n^2}}$, ${\displaystyle \cos \beta =\frac{m^2-n^2}{m^2+n^2}}$, and ${\displaystyle \tan \beta =\frac{2mn}{m^2-n^2}}$.

A variant

The following variant of Euclid's formula is sometimes more convenient, as being more symmetric in m and n (same parity condition on m and n).

If m and n are two odd integers such that m > n, then

${\displaystyle a=mn,b=\frac{m^2-n^2}{2},c=\frac{m^2+n^2}{2}}$

are three integers that form a Pythagorean triple, which is primitive if and only if m and n are coprime. Conversely, every primitive Pythagorean triple arises (after the exchange of a and b, if a is even) from a unique pair m > n > 0 of coprime odd integers.

Elementary properties of primitive Pythagorean triples

General properties

The properties of a primitive Pythagorean triple (a, b, c) with a < b < c (without specifying which of a or b is even and which is odd) include:

• $\frac{(c-a)(c-b)}{2}$ is always a perfect square. As it is only a necessary condition but not a sufficient one, it can be used in checking if a given triple of numbers is not a Pythagorean triple when they fail the test. For example, the triples {6, 12, 18} and {1, 8, 9} each pass the test that (c − a)(c − b)/2 is a perfect square, but neither is a Pythagorean triple.
• When a triple of numbers a, b and c forms a primitive Pythagorean triple, then (c minus the even leg) and one-half of (c minus the odd leg) are both perfect squares; however this is not a sufficient condition, as the numbers {1, 8, 9} pass the perfect squares test but are not a Pythagorean triple since 1² + 8² ≠ 9².
• At most one of a, b, c is a square.
• The area of a Pythagorean triangle cannot be the square or twice the square of a natural number.
• Exactly one of a, b is divisible by 2 (is even), but never c.
• Exactly one of a, b is divisible by 3, but never c.
• Exactly one of a, b is divisible by 4, but never c (because c is never even).
• Exactly one of a, b, c is divisible by 5.
• The largest number that always divides abc is 60.
• Any odd number of the form 2m+1, where m is an integer and m>1, can be the odd leg of a primitive Pythagorean triple [PPT]. See almost-isosceles PPT section below. However, only even numbers divisible by 4 can be the even leg of a PPT. This is because Euclid's formula for the even leg given above is 2mn and one of m or n must be even.
• The hypotenuse c is the sum of two squares. This requires all of its prime factors to be primes of the form 4n + 1. Therefore, c is of the form 4n + 1. A sequence of possible hypotenuse numbers for a PPT can be found at (sequence A008846 in the OEIS).
• The area (K = ab/2) is a congruent number divisible by 6.
• In every Pythagorean triangle, the radius of the incircle and the radii of the three excircles are natural numbers. Specifically, for a primitive triple the radius of the incircle is r = n(m − n), and the radii of the excircles opposite the sides m² − n², 2mn, and the hypotenuse m² + n² are respectively m(m − n), n(m + n), and m(m + n).
• As for any right triangle, the converse of Thales' theorem says that the diameter of the circumcircle equals the hypotenuse; hence for primitive triples the circumdiameter is m² + n², and the circumradius is half of this and thus is rational but non-integer (since m and n have opposite parity).
• When the area of a Pythagorean triangle is multiplied by the curvatures of its incircle and 3 excircles, the result is four positive integers w > x > y > z, respectively. Integers −w, x, y, z satisfy Descartes's Circle Equation. Equivalently, the radius of the outer Soddy circle of any right triangle is equal to its semiperimeter. The outer Soddy center is located at D, where ACBD is a rectangle, ACB the right triangle and AB its hypotenuse.
• Only two sides of a primitive Pythagorean triple can be simultaneously prime because by Euclid's formula for generating a primitive Pythagorean triple, one of the legs must be composite and even. However, only one side can be an integer of perfect power $p\ge 2$ because if two sides were integers of perfect powers with equal exponent $p$ it would contradict the fact that there are no integer solutions to the Diophantine equation ${\displaystyle x^{2p}\pm y^{2p}=z^2}$, with $x$, $y$ and $z$ being pairwise coprime.
• There are no Pythagorean triangles in which the hypotenuse and one leg are the legs of another Pythagorean triangle; this is one of the equivalent forms of Fermat's right triangle theorem.
• Each primitive Pythagorean triangle has a ratio of area, K, to squared semiperimeter, s, that is unique to itself and is given by

$\frac{K}{s^2}=\frac{n(m-n)}{m(m+n)}=1-\frac{c}{s}.$

• No primitive Pythagorean triangle has an integer altitude from the hypotenuse; that is, every primitive Pythagorean triangle is indecomposable.
• The set of all primitive Pythagorean triples forms a rooted ternary tree in a natural way; see Tree of primitive Pythagorean triples.
• Neither of the acute angles of a Pythagorean triangle can be a rational number of degrees. (This follows from Niven's theorem.)

Special cases

In addition, special Pythagorean triples with certain additional properties can be guaranteed to exist:

• Every integer greater than 2 that is not congruent to 2 mod 4 (in other words, every integer greater than 2 which is not of the form 4k + 2) is part of a primitive Pythagorean triple. (If the integer has the form 4k, one may take n = 1 and m = 2k in Euclid's formula; if the integer is 2k + 1, one may take n = k and m = k + 1.)
• Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple. For example, the integers 6, 10, 14, and 18 are not part of primitive triples, but are part of the non-primitive triples (6, 8, 10), (14, 48, 50) and (18, 80, 82).
• There exist infinitely many Pythagorean triples in which the hypotenuse and the longest leg differ by exactly one. Such triples are necessarily primitive and have the form (2n + 1, 2n² + 2n, 2n² + 2n +1). This results from Euclid's formula by remarking that the condition implies that the triple is primitive and must verify (m² + n²) - 2mn = 1. This implies (m – n)² = 1, and thus m = n + 1. The above form of the triples results thus of substituting m for n + 1 in Euclid's formula.
• There exist infinitely many primitive Pythagorean triples in which the hypotenuse and the longest leg differ by exactly two. They are all primitive, and are obtained by putting n = 1 in Euclid's formula. More generally, for every integer k > 0, there exist infinitely many primitive Pythagorean triples in which the hypotenuse and the odd leg differ by 2k². They are obtained by putting n = k in Euclid's formula.
• There exist infinitely many Pythagorean triples in which the two legs differ by exactly one. For example, 20² + 21² = 29²; these are generated by Euclid's formula when ${\displaystyle \frac{m-n}{n}}$ is a convergent to $\sqrt{2}$.
• For each natural number k, there exist k Pythagorean triples with different hypotenuses and the same area.
• For each natural number k, there exist at least k different primitive Pythagorean triples with the same leg a, where a is some natural number (the length of the even leg is 2mn, and it suffices to choose a with many factorizations, for example a = 4b, where b is a product of k different odd primes; this produces at least 2^k different primitive triples).
• For each natural number k, there exist at least k different Pythagorean triples with the same hypotenuse.
• If c = p^e is a prime power, there exists a primitive Pythagorean triple a² + b² = c² if and only if the prime p has the form 4n + 1; this triple is unique up to the exchange of a and b.
• More generally, a positive integer c is the hypotenuse of a primitive Pythagorean triple if and only if each prime factor of c is congruent to 1 modulo 4; that is, each prime factor has the form 4n + 1. In this case, the number of primitive Pythagorean triples (a, b, c) with a < b is 2^k−1, where k is the number of distinct prime factors of c.
• There exist infinitely many Pythagorean triples with square numbers for both the hypotenuse c and the sum of the legs a + b. According to Fermat, the smallest such triple has sides a = 4,565,486,027,761; b = 1,061,652,293,520; and c = 4,687,298,610,289. Here a + b = 2,372,159² and c = 2,165,017². This is generated by Euclid's formula with parameter values m = 2,150,905 and n = 246,792.
• There exist non-primitive Pythagorean triangles with integer altitude from the hypotenuse. Such Pythagorean triangles are known as decomposable since they can be split along this altitude into two separate and smaller Pythagorean triangles.

Geometry of Euclid's formula

Rational points on a unit circle

3,4,5 maps to x,y point (4/5,3/5) on the unit circle

The rational points on a circle correspond, under stereographic projection, to the rational points of the line.

Euclid's formula for a Pythagorean triple

${\displaystyle a=m^2-n^2,b=2mn,c=m^2+n^2}$

can be understood in terms of the geometry of rational points on the unit circle (Trautman 1998).

In fact, a point in the Cartesian plane with coordinates (x, y) belongs to the unit circle if x² + y² = 1. The point is rational if x and y are rational numbers, that is, if there are coprime integers a, b, c such that

${\displaystyle (\frac{a}{c}{)}^2+(\frac{b}{c}{)}^2=1.}$

By multiplying both members by c², one can see that the rational points on the circle are in one-to-one correspondence with the primitive Pythagorean triples.

The unit circle may also be defined by a parametric equation

${\displaystyle x=\frac{1-t^2}{1+t^2}y=\frac{2t}{1+t^2}.}$

Euclid's formula for Pythagorean triples and the inverse relationship t = y / (x + 1) mean that, except for (−1, 0), a point (x, y) on the circle is rational if and only if the corresponding value of t is a rational number. Note that t = y / (x + 1) = b / (a + c) = n / m is also the tangent of half of the angle that is opposite the triangle side of length b.

Stereographic approach

Stereographic projection of the unit circle onto the x-axis. Given a point P on the unit circle, draw a line from P to the point N = (0, 1) (the north pole). The point P′ where the line intersects the x-axis is the stereographic projection of P. Inversely, starting with a point P′ on the x-axis, and drawing a line from P′ to N, the inverse stereographic projection is the point P where the line intersects the unit circle.

There is a correspondence between points on the unit circle with rational coordinates and primitive Pythagorean triples. At this point, Euclid's formulae can be derived either by methods of trigonometry or equivalently by using the stereographic projection.

For the stereographic approach, suppose that P′ is a point on the x-axis with rational coordinates

$P^{\prime }=\left(\frac{m}{n},0\right).$

Then, it can be shown by basic algebra that the point P has coordinates

$P=\left(\frac{2\left(\frac{m}{n}\right)}{{\left(\frac{m}{n}\right)}^2+1},\frac{{\left(\frac{m}{n}\right)}^2-1}{{\left(\frac{m}{n}\right)}^2+1}\right)=\left(\frac{2mn}{m^2+n^2},\frac{m^2-n^2}{m^2+n^2}\right).$

This establishes that each rational point of the x-axis goes over to a rational point of the unit circle. The converse, that every rational point of the unit circle comes from such a point of the x-axis, follows by applying the inverse stereographic projection. Suppose that P(x, y) is a point of the unit circle with x and y rational numbers. Then the point P′ obtained by stereographic projection onto the x-axis has coordinates

$\left(\frac{x}{1-y},0\right)$

which is rational.

In terms of algebraic geometry, the algebraic variety of rational points on the unit circle is birational to the affine line over the rational numbers. The unit circle is thus called a rational curve, and it is this fact which enables an explicit parameterization of the (rational number) points on it by means of rational functions.

Pythagorean triangles in a 2D lattice

A 2D lattice is a regular array of isolated points where if any one point is chosen as the Cartesian origin (0, 0), then all the other points are at (x, y) where x and y range over all positive and negative integers. Any Pythagorean triangle with triple (a, b, c) can be drawn within a 2D lattice with vertices at coordinates (0, 0), (a, 0) and (0, b). The count of lattice points lying strictly within the bounds of the triangle is given by   $\frac{(a-1)(b-1)-gcd(a,b)+1}{2};$ for primitive Pythagorean triples this interior lattice count is  $\frac{(a-1)(b-1)}{2}.$ The area (by Pick's theorem equal to one less than the interior lattice count plus half the boundary lattice count) equals  $\frac{ab}{2}$ .

The first occurrence of two primitive Pythagorean triples sharing the same area occurs with triangles with sides (20, 21, 29), (12, 35, 37) and common area 210 (sequence A093536 in the OEIS). The first occurrence of two primitive Pythagorean triples sharing the same interior lattice count occurs with (18108, 252685, 253333), (28077, 162964, 165365) and interior lattice count 2287674594 (sequence A225760 in the OEIS). Three primitive Pythagorean triples have been found sharing the same area: (4485, 5852, 7373), (3059, 8580, 9109), (1380, 19019, 19069) with area 13123110. As yet, no set of three primitive Pythagorean triples have been found sharing the same interior lattice count.

Enumeration of primitive Pythagorean triples

By Euclid's formula all primitive Pythagorean triples can be generated from integers $m$ and $n$ with ${\displaystyle m>n>0}$, $m+n$ odd and ${\displaystyle gcd(m,n)=1}$. Hence there is a 1 to 1 mapping of rationals (in lowest terms) to primitive Pythagorean triples where $\frac{n}{m}$ is in the interval $(0,1)$ and $m+n$ odd.

The reverse mapping from a primitive triple ${\displaystyle (a,b,c)}$ where ${\displaystyle c>b>a>0}$ to a rational $\frac{n}{m}$ is achieved by studying the two sums $a+c$ and $b+c$. One of these sums will be a square that can be equated to ${\displaystyle (m+n{)}^2}$ and the other will be twice a square that can be equated to ${\displaystyle 2m^2}$. It is then possible to determine the rational $\frac{n}{m}$.

In order to enumerate primitive Pythagorean triples the rational can be expressed as an ordered pair $(n,m)$ and mapped to an integer using a pairing function such as Cantor's pairing function. An example can be seen at (sequence A277557 in the OEIS). It begins

${\displaystyle 8,18,19,32,33,34,\dots }$ and gives rationals

${\displaystyle \frac{1}{2},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{2}{5},\frac{1}{6},\dots }$ these, in turn, generate primitive triples

${\displaystyle (3,4,5),(5,12,13),(8,15,17),(7,24,25),(20,21,29),(12,35,37),\dots }$

Spinors and the modular group

Pythagorean triples can likewise be encoded into a square matrix of the form

$X=\left[\begin{array}{cc}c+b& a\\ a& c-b\end{array}\right].$

A matrix of this form is symmetric. Furthermore, the determinant of X is

$detX=c^2-a^2-b^2$

which is zero precisely when (a,b,c) is a Pythagorean triple. If X corresponds to a Pythagorean triple, then as a matrix it must have rank 1.

Since X is symmetric, it follows from a result in linear algebra that there is a column vector ξ = [m n]^T such that the outer product

$X=2\left[\begin{array}{c}m\\ n\end{array}\right][mn]=2\xi {\xi }^T$

(1)

holds, where the T denotes the matrix transpose. The vector ξ is called a spinor (for the Lorentz group SO(1, 2)). In abstract terms, the Euclid formula means that each primitive Pythagorean triple can be written as the outer product with itself of a spinor with integer entries.

The modular group Γ is the set of 2×2 matrices with integer entries

$A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right]$

with determinant equal to one: αδ − βγ = 1. This set forms a group, since the inverse of a matrix in Γ is again in Γ, as is the product of two matrices in Γ. The modular group acts on the collection of all integer spinors. Furthermore, the group is transitive on the collection of integer spinors with relatively prime entries. For if [m n]^T has relatively prime entries, then

$\left[\begin{array}{cc}m& -v\\ n& u\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}m\\ n\end{array}\right]$

where u and v are selected (by the Euclidean algorithm) so that mu + nv = 1.

By acting on the spinor ξ , the action of Γ goes over to an action on Pythagorean triples, provided one allows for triples with possibly negative components. Thus if A is a matrix in Γ, then

$2(A\xi )(A\xi {)}^T=AX{A}^T$

(2)

gives rise to an action on the matrix X in (1). This does not give a well-defined action on primitive triples, since it may take a primitive triple to an imprimitive one. It is convenient at this point (per Trautman 1998) to call a triple (a,b,c) standard if c > 0 and either (a,b,c) are relatively prime or (a/2,b/2,c/2) are relatively prime with a/2 odd. If the spinor [m n]^T has relatively prime entries, then the associated triple (a,b,c) determined by (1) is a standard triple. It follows that the action of the modular group is transitive on the set of standard triples.

Alternatively, restrict attention to those values of m and n for which m is odd and n is even. Let the subgroup Γ(2) of Γ be the kernel of the group homomorphism

$\mathrm{\Gamma }=\mathrm{S}\mathrm{L}(2,\mathbf{Z})\to \mathrm{S}\mathrm{L}(2,{\mathbf{Z}}_2)$

where SL(2,Z₂) is the special linear group over the finite field Z₂ of integers modulo 2. Then Γ(2) is the group of unimodular transformations which preserve the parity of each entry. Thus if the first entry of ξ is odd and the second entry is even, then the same is true of Aξ for all A ∈ Γ(2). In fact, under the action (2), the group Γ(2) acts transitively on the collection of primitive Pythagorean triples (Alperin 2005).

The group Γ(2) is the free group whose generators are the matrices

$U=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right],L=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right].$

Consequently, every primitive Pythagorean triple can be obtained in a unique way as a product of copies of the matrices U and L.

Parent/child relationships

Main article: Tree of Pythagorean triples

By a result of Berggren (1934), all primitive Pythagorean triples can be generated from the (3, 4, 5) triangle by using the three linear transformations T₁, T₂, T₃ below, where a, b, c are sides of a triple:

	new side a	new side b	new side c
T1:	a − 2b + 2c	2a − b + 2c	2a − 2b + 3c
T2:	a + 2b + 2c	2a + b + 2c	2a + 2b + 3c
T3:	−a + 2b + 2c	−2a + b + 2c	−2a + 2b + 3c

In other words, every primitive triple will be a "parent" to three additional primitive triples. Starting from the initial node with a = 3, b = 4, and c = 5, the operation T₁ produces the new triple

(3 − (2×4) + (2×5), (2×3) − 4 + (2×5), (2×3) − (2×4) + (3×5)) = (5, 12, 13),

and similarly T₂ and T₃ produce the triples (21, 20, 29) and (15, 8, 17).

The linear transformations T₁, T₂, and T₃ have a geometric interpretation in the language of quadratic forms. They are closely related to (but are not equal to) reflections generating the orthogonal group of x² + y² − z² over the integers.

Relation to Gaussian integers

Alternatively, Euclid's formulae can be analyzed and proved using the Gaussian integers. Gaussian integers are complex numbers of the form α = u + vi, where u and v are ordinary integers and i is the square root of negative one. The units of Gaussian integers are ±1 and ±i. The ordinary integers are called the rational integers and denoted as 'Z'. The Gaussian integers are denoted as Z[i]. The right-hand side of the Pythagorean theorem may be factored in Gaussian integers:

$c^2=a^2+b^2=(a+bi)\overline{(a+bi)}=(a+bi)(a-bi).$

A primitive Pythagorean triple is one in which a and b are coprime, i.e., they share no prime factors in the integers. For such a triple, either a or b is even, and the other is odd; from this, it follows that c is also odd.

The two factors z := a + bi and z* := a − bi of a primitive Pythagorean triple each equal the square of a Gaussian integer. This can be proved using the property that every Gaussian integer can be factored uniquely into Gaussian primes up to units. (This unique factorization follows from the fact that, roughly speaking, a version of the Euclidean algorithm can be defined on them.) The proof has three steps. First, if a and b share no prime factors in the integers, then they also share no prime factors in the Gaussian integers. (Assume a = gu and b = gv with Gaussian integers g, u and v and g not a unit. Then u and v lie on the same line through the origin. All Gaussian integers on such a line are integer multiples of some Gaussian integer h. But then the integer gh ≠ ±1 divides both a and b.) Second, it follows that z and z* likewise share no prime factors in the Gaussian integers. For if they did, then their common divisor δ would also divide z + z* = 2a and z − z* = 2ib. Since a and b are coprime, that implies that δ divides 2 = (1 + i)(1 − i) = i(1 − i)². From the formula c² = zz*, that in turn would imply that c is even, contrary to the hypothesis of a primitive Pythagorean triple. Third, since c² is a square, every Gaussian prime in its factorization is doubled, i.e., appears an even number of times. Since z and z* share no prime factors, this doubling is also true for them. Hence, z and z* are squares.

Thus, the first factor can be written

$a+bi=\epsilon {\left(m+ni\right)}^2,\epsilon \in \{\pm 1,\pm i\}.$

The real and imaginary parts of this equation give the two formulas:

$\{\begin{array}{ll}\epsilon =+1,& a=+\left(m^2-n^2\right),b=+2mn;\\ \epsilon =-1,& a=-\left(m^2-n^2\right),b=-2mn;\\ \epsilon =+i,& a=-2mn,b=+\left(m^2-n^2\right);\\ \epsilon =-i,& a=+2mn,b=-\left(m^2-n^2\right).\end{array}$

For any primitive Pythagorean triple, there must be integers m and n such that these two equations are satisfied. Hence, every Pythagorean triple can be generated from some choice of these integers.

As perfect square Gaussian integers

If we consider the square of a Gaussian integer we get the following direct interpretation of Euclid's formula as representing the perfect square of a Gaussian integer.

$(m+ni{)}^2=(m^2-n^2)+2mni.$

Using the facts that the Gaussian integers are a Euclidean domain and that for a Gaussian integer p $|p{|}^2$ is always a square it is possible to show that a Pythagorean triple corresponds to the square of a prime Gaussian integer if the hypotenuse is prime.

If the Gaussian integer is not prime then it is the product of two Gaussian integers p and q with $|p{|}^2$ and $|q{|}^2$ integers. Since magnitudes multiply in the Gaussian integers, the product must be $|p||q|$, which when squared to find a Pythagorean triple must be composite. The contrapositive completes the proof.

Distribution of triples

A scatter plot of the legs (a,b) of the first Pythagorean triples with a and b less than 4500.

There are a number of results on the distribution of Pythagorean triples. In the scatter plot, a number of obvious patterns are already apparent. Whenever the legs (a,b) of a primitive triple appear in the plot, all integer multiples of (a,b) must also appear in the plot, and this property produces the appearance of lines radiating from the origin in the diagram.

Within the scatter, there are sets of parabolic patterns with a high density of points and all their foci at the origin, opening up in all four directions. Different parabolas intersect at the axes and appear to reflect off the axis with an incidence angle of 45 degrees, with a third parabola entering in a perpendicular fashion. Within this quadrant, each arc centered on the origin shows that section of the parabola that lies between its tip and its intersection with its semi-latus rectum.

These patterns can be explained as follows. If $a^2/4n$ is an integer, then (a, $|n-a^2/4n|$, $n+a^2/4n$) is a Pythagorean triple. (In fact every Pythagorean triple (a, b, c) can be written in this way with integer n, possibly after exchanging a and b, since $n=(b+c)/2$ and a and b cannot both be odd.) The Pythagorean triples thus lie on curves given by $b=|n-a^2/4n|$, that is, parabolas reflected at the a-axis, and the corresponding curves with a and b interchanged. If a is varied for a given n (i.e. on a given parabola), integer values of b occur relatively frequently if n is a square or a small multiple of a square. If several such values happen to lie close together, the corresponding parabolas approximately coincide, and the triples cluster in a narrow parabolic strip. For instance, 38² = 1444, 2 × 27² = 1458, 3 × 22² = 1452, 5 × 17² = 1445 and 10 × 12² = 1440; the corresponding parabolic strip around n ≈ 1450 is clearly visible in the scatter plot.

The angular properties described above follow immediately from the functional form of the parabolas. The parabolas are reflected at the a-axis at a = 2n, and the derivative of b with respect to a at this point is –1; hence the incidence angle is 45°. Since the clusters, like all triples, are repeated at integer multiples, the value 2n also corresponds to a cluster. The corresponding parabola intersects the b-axis at right angles at b = 2n, and hence its reflection upon interchange of a and b intersects the a-axis at right angles at a = 2n, precisely where the parabola for n is reflected at the a-axis. (The same is of course true for a and b interchanged.)

Albert Fässler and others provide insights into the significance of these parabolas in the context of conformal mappings.

Special cases and related equations

The Platonic sequence

The case n = 1 of the more general construction of Pythagorean triples has been known for a long time. Proclus, in his commentary to the 47th Proposition of the first book of Euclid's Elements, describes it as follows:

Certain methods for the discovery of triangles of this kind are handed down, one which they refer to Plato, and another to Pythagoras. (The latter) starts from odd numbers. For it makes the odd number the smaller of the sides about the right angle; then it takes the square of it, subtracts unity and makes half the difference the greater of the sides about the right angle; lastly it adds unity to this and so forms the remaining side, the hypotenuse.
...For the method of Plato argues from even numbers. It takes the given even number and makes it one of the sides about the right angle; then, bisecting this number and squaring the half, it adds unity to the square to form the hypotenuse, and subtracts unity from the square to form the other side about the right angle. ... Thus it has formed the same triangle that which was obtained by the other method.

In equation form, this becomes:

a is odd (Pythagoras, c. 540 BC):

$\text{side }a:\text{side }b=\frac{a^2-1}{2}:\text{side }c=\frac{a^2+1}{2}.$

a is even (Plato, c. 380 BC):

$\text{side }a:\text{side }b={\left(\frac{a}{2}\right)}^2-1:\text{side }c={\left(\frac{a}{2}\right)}^2+1$

It can be shown that all Pythagorean triples can be obtained, with appropriate rescaling, from the basic Platonic sequence (a, (a² − 1)/2 and (a² + 1)/2) by allowing a to take non-integer rational values. If a is replaced with the fraction m/n in the sequence, the result is equal to the 'standard' triple generator (2mn, m² − n²,m² + n²) after rescaling. It follows that every triple has a corresponding rational a value which can be used to generate a similar triangle (one with the same three angles and with sides in the same proportions as the original). For example, the Platonic equivalent of (56, 33, 65) is generated by a = m/n = 7/4 as (a, (a² –1)/2, (a²+1)/2) = (56/32, 33/32, 65/32). The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII.

The Jacobi–Madden equation

Main article: Jacobi–Madden equation

The equation,

$a^4+b^4+c^4+d^4=(a+b+c+d{)}^4$

is equivalent to the special Pythagorean triple,

$(a^2+ab+b^2{)}^2+(c^2+cd+d^2{)}^2=((a+b{)}^2+(a+b)(c+d)+(c+d{)}^2{)}^2$

There is an infinite number of solutions to this equation as solving for the variables involves an elliptic curve. Small ones are,

$a,b,c,d=-2634,955,1770,5400$

$a,b,c,d=-31764,7590,27385,48150$

Equal sums of two squares

One way to generate solutions to $a^2+b^2=c^2+d^2$ is to parametrize a, b, c, d in terms of integers m, n, p, q as follows:

$(m^2+n^2)(p^2+q^2)=(mp-nq{)}^2+(np+mq{)}^2=(mp+nq{)}^2+(np-mq{)}^2.$

Equal sums of two fourth powers

Given two sets of Pythagorean triples,

$(a^2-b^2{)}^2+(2ab{)}^2=(a^2+b^2{)}^2$

$(c^2-d^2{)}^2+(2cd{)}^2=(c^2+d^2{)}^2$

the problem of finding equal products of a non-hypotenuse side and the hypotenuse,

$(a^2-b^2)(a^2+b^2)=(c^2-d^2)(c^2+d^2)$

is easily seen to be equivalent to the equation,

$a^4-b^4=c^4-d^4$

and was first solved by Euler as $a,b,c,d=133,59,158,134$. Since he showed this is a rational point in an elliptic curve, then there is an infinite number of solutions. In fact, he also found a 7th degree polynomial parameterization.

Descartes' Circle Theorem

For the case of Descartes' circle theorem where all variables are squares,

$2(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2{)}^2$

Euler showed this is equivalent to three simultaneous Pythagorean triples,

$(2ab{)}^2+(2cd{)}^2=(a^2+b^2-c^2-d^2{)}^2$

$(2ac{)}^2+(2bd{)}^2=(a^2-b^2+c^2-d^2{)}^2$

$(2ad{)}^2+(2bc{)}^2=(a^2-b^2-c^2+d^2{)}^2$

There is also an infinite number of solutions, and for the special case when $a+b=c$, then the equation simplifies to,

$4(a^2+ab+b^2)=d^2$

with small solutions as $a,b,c,d=3,5,8,14$ and can be solved as binary quadratic forms.

Almost-isosceles Pythagorean triples

No Pythagorean triples are isosceles, because the ratio of the hypotenuse to either other side is √2, but √2 cannot be expressed as the ratio of 2 integers.

There are, however, right-angled triangles with integral sides for which the lengths of the non-hypotenuse sides differ by one, such as,

$3^2+4^2=5^2$

${20}^2+{21}^2={29}^2$

and an infinite number of others. They can be completely parameterized as,

${\displaystyle {\left(\frac{x-1}{2}\right)}^2+{\left(\frac{x+1}{2}\right)}^2=y^2}$

where {x, y} are the solutions to the Pell equation $x^2-2y^2=-1$.

If a, b, c are the sides of this type of primitive Pythagorean triple (PPT) then the solution to the Pell equation is given by the recursive formula

${\displaystyle a_n=6a_{n-1}-a_{n-2}+2}$ with ${\displaystyle a_1=3}$ and ${\displaystyle a_2=20}$

${\displaystyle b_n=6b_{n-1}-b_{n-2}-2}$ with ${\displaystyle b_1=4}$ and ${\displaystyle b_2=21}$

${\displaystyle c_n=6c_{n-1}-c_{n-2}}$ with ${\displaystyle c_1=5}$ and ${\displaystyle c_2=29}$.

This sequence of PPTs forms the central stem (trunk) of the rooted ternary tree of PPTs.

When it is the longer non-hypotenuse side and hypotenuse that differ by one, such as in

$5^2+{12}^2={13}^2$

$7^2+{24}^2={25}^2$

then the complete solution for the PPT a, b, c is

${\displaystyle a=2m+1,b=2m^2+2m,c=2m^2+2m+1}$

and

${\displaystyle (2m+1{)}^2+(2m^2+2m{)}^2=(2m^2+2m+1{)}^2}$

where integer $m>0$ is the generating parameter.

It shows that all odd numbers (greater than 1) appear in this type of almost-isosceles PPT. This sequence of PPTs forms the right hand side outer stem of the rooted ternary tree of PPTs.

Another property of this type of almost-isosceles PPT is that the sides are related such that

${\displaystyle a^b+b^a=Kc}$

for some integer $K$. Or in other words ${\displaystyle a^b+b^a}$ is divisible by $c$ such as in

${\displaystyle (5^{12}+{12}^5)/13=18799189}$.

Fibonacci numbers in Pythagorean triples

Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple, obtained from the formula

$${\displaystyle (F_n{F}_{n+3}{)}^2+(2F_{n+1}{F}_{n+2}{)}^2=F_{2n+3}^2.}$$

The sequence of Pythagorean triangles obtained from this formula has sides of lengths

(3,4,5), (5,12,13), (16,30,34), (39,80,89), ...

The middle side of each of these triangles is the sum of the three sides of the preceding triangle.

Generalizations

There are several ways to generalize the concept of Pythagorean triples.

Pythagorean n-tuple

See also: Pythagorean quadruple

The expression

${\displaystyle {\left(m_1^2-m_2^2-\dots -m_n^2\right)}^2+\sum _{k=2}^n(2m_1{m}_k{)}^2={\left(m_1^2+\dots +m_n^2\right)}^2}$

is a Pythagorean n-tuple for any tuple of positive integers (m₁, ..., m_n) with m²
₁ > m²
₂ + ... + m²
_n. The Pythagorean n-tuple can be made primitive by dividing out by the largest common divisor of its values.

Furthermore, any primitive Pythagorean n-tuple a²
₁ + ... + a²
_n = c² can be found by this approach. Use (m₁, ..., m_n) = (c + a₁, a₂, ..., a_n) to get a Pythagorean n-tuple by the above formula and divide out by the largest common integer divisor, which is 2m₁ = 2(c + a₁). Dividing out by the largest common divisor of these (m₁, ..., m_n) values gives the same primitive Pythagorean n-tuple; and there is a one-to-one correspondence between tuples of setwise coprime positive integers (m₁, ..., m_n) satisfying m²
₁ > m²
₂ + ... + m²
_n and primitive Pythagorean n-tuples.

Examples of the relationship between setwise coprime values $\overrightarrow{m}$ and primitive Pythagorean n-tuples include:

${\displaystyle \begin{array}{rl}\overrightarrow{m}=(1)& \leftrightarrow 1^2=1^2\\ \overrightarrow{m}=(2,1)& \leftrightarrow 3^2+4^2=5^2\\ \overrightarrow{m}=(2,1,1)& \leftrightarrow 1^2+2^2+2^2=3^2\\ \overrightarrow{m}=(3,1,1,1)& \leftrightarrow 1^2+1^2+1^2+1^2=2^2\\ \overrightarrow{m}=(5,1,1,2,3)& \leftrightarrow 1^2+1^2+1^2+2^2+3^2=4^2\\ \overrightarrow{m}=(4,1,1,1,1,2)& \leftrightarrow 1^2+1^2+1^2+1^2+1^2+2^2=3^2\\ \overrightarrow{m}=(5,1,1,1,2,2,2)& \leftrightarrow 1^2+1^2+1^2+1^2+2^2+2^2+2^2=4^2\end{array}}$

Consecutive squares

Since the sum F(k,m) of k consecutive squares beginning with m² is given by the formula,

$F(k,m)=km(k-1+m)+\frac{k(k-1)(2k-1)}{6}$

one may find values (k, m) so that F(k,m) is a square, such as one by Hirschhorn where the number of terms is itself a square,

$m=\frac{v^4-24v^2-25}{48},k=v^2,F(m,k)=\frac{v^5+47v}{48}$

and v ≥ 5 is any integer not divisible by 2 or 3. For the smallest case v = 5, hence k = 25, this yields the well-known cannonball-stacking problem of Lucas,

$0^2+1^2+2^2+\cdots +{24}^2={70}^2$

a fact which is connected to the Leech lattice.

In addition, if in a Pythagorean n-tuple (n ≥ 4) all addends are consecutive except one, one can use the equation,

$F(k,m)+p^2=(p+1{)}^2$

Since the second power of p cancels out, this is only linear and easily solved for as $p=\frac{F(k,m)-1}{2}$ though k, m should be chosen so that p is an integer, with a small example being k = 5, m = 1 yielding,

$1^2+2^2+3^2+4^2+5^2+{27}^2={28}^2$

Thus, one way of generating Pythagorean n-tuples is by using, for various x,

$x^2+(x+1{)}^2+\cdots +(x+q{)}^2+p^2=(p+1{)}^2,$

where q = n–2 and where

$p=\frac{(q+1)x^2+q(q+1)x+\frac{q(q+1)(2q+1)}{6}-1}{2}.$

Fermat's Last Theorem

Main article: Fermat's Last Theorem

A generalization of the concept of Pythagorean triples is the search for triples of positive integers a, b, and c, such that aⁿ + bⁿ = cⁿ, for some n strictly greater than 2. Pierre de Fermat in 1637 claimed that no such triple exists, a claim that came to be known as Fermat's Last Theorem because it took longer than any other conjecture by Fermat to be proved or disproved. The first proof was given by Andrew Wiles in 1994.

n − 1 or n nth powers summing to an nth power

Main article: Euler's sum of powers conjecture

Another generalization is searching for sequences of n + 1 positive integers for which the nth power of the last is the sum of the nth powers of the previous terms. The smallest sequences for known values of n are:

• n = 3: {3, 4, 5; 6}.
• n = 4: {30, 120, 272, 315; 353}
• n = 5: {19, 43, 46, 47, 67; 72}
• n = 7: {127, 258, 266, 413, 430, 439, 525; 568}
• n = 8: {90, 223, 478, 524, 748, 1088, 1190, 1324; 1409}

For the n = 3 case, in which $x^3+y^3+z^3=w^3,$ called the Fermat cubic, a general formula exists giving all solutions.

A slightly different generalization allows the sum of (k + 1) nth powers to equal the sum of (n − k) nth powers. For example:

• (n = 3): 1³ + 12³ = 9³ + 10³, made famous by Hardy's recollection of a conversation with Ramanujan about the number 1729 being the smallest number that can be expressed as a sum of two cubes in two distinct ways.

There can also exist n − 1 positive integers whose nth powers sum to an nth power (though, by Fermat's Last Theorem, not for n = 3); these are counterexamples to Euler's sum of powers conjecture. The smallest known counterexamples are

• n = 4: (95800, 217519, 414560; 422481)
• n = 5: (27, 84, 110, 133; 144)

Heronian triangle triples

Main article: Heronian triangle

A Heronian triangle is commonly defined as one with integer sides whose area is also an integer. The lengths of the sides of such a triangle form a Heronian triple (a, b, c) for a ≤ b ≤ c. Every Pythagorean triple is a Heronian triple, because at least one of the legs a, b must be even in a Pythagorean triple, so the area ab/2 is an integer. Not every Heronian triple is a Pythagorean triple, however, as the example (4, 13, 15) with area 24 shows.

If (a, b, c) is a Heronian triple, so is (ka, kb, kc) where k is any positive integer; its area will be the integer that is k² times the integer area of the (a, b, c) triangle. The Heronian triple (a, b, c) is primitive provided a, b, c are setwise coprime. (With primitive Pythagorean triples the stronger statement that they are pairwise coprime also applies, but with primitive Heronian triangles the stronger statement does not always hold true, such as with (7, 15, 20).) Here are a few of the simplest primitive Heronian triples that are not Pythagorean triples:

(4, 13, 15) with area 24

(3, 25, 26) with area 36

(7, 15, 20) with area 42

(6, 25, 29) with area 60

(11, 13, 20) with area 66

(13, 14, 15) with area 84

(13, 20, 21) with area 126

By Heron's formula, the extra condition for a triple of positive integers (a, b, c) with a < b < c to be Heronian is that

(a² + b² + c²)² − 2(a⁴ + b⁴ + c⁴)

or equivalently

2(a²b² + a²c² + b²c²) − (a⁴ + b⁴ + c⁴)

be a nonzero perfect square divisible by 16.

Application to cryptography

Primitive Pythagorean triples have been used in cryptography as random sequences and for the generation of keys.

The Twilight Zone (1959 TV series)

From Wikipedia, the free encyclopedia

https://en.wikipedia.org/wiki/The_Twilight_Zone_(1959_TV_series) 

 

The Twilight Zone
Genre	Anthology Horror Science fiction
Created by	Rod Serling
Presented by	Rod Serling
Composers	Bernard Herrmann (also season 1 theme) Marius Constant (theme from season 2 forward) Jerry Goldsmith Fred Steiner Leith Stevens Leonard Rosenman Franz Waxman
Country of origin	United States
No. of seasons	5
No. of episodes	156 (list of episodes)
Production
Executive producer	Rod Serling
Producers	Buck Houghton (1959–62) Herbert Hirschman (1963) Bert Granet (1963–64) William Froug (1963–64)
Cinematography	George T. Clemens
Running time	25 min. (seasons 1–3, 5) 51 min. (season 4)[citation needed]
Production companies	Cayuga Productions, Inc. CBS Productions
Release
Original network	CBS
Audio format	Mono
Original release	October 2, 1959 – June 19, 1964
Related
The Twilight Zone (1985–89) The Twilight Zone (2002–03) The Twilight Zone (2019–20)

The Twilight Zone (marketed as Twilight Zone for its final two seasons) is an American science fiction horror anthology television series created and presented by Rod Serling, which ran for five seasons on CBS from October 2, 1959, to June 19, 1964. Each episode presents a stand-alone story in which characters find themselves dealing with often disturbing or unusual events, an experience described as entering "the Twilight Zone", often with a surprise ending and a moral. Although predominantly science-fiction, the show's paranormal and Kafkaesque events leaned the show towards fantasy and horror. The phrase "twilight zone", inspired by the series, is used to describe surreal experiences.

The series featured both established stars and younger actors who would become much better known later. Serling served as executive producer and head writer; he wrote or co-wrote 92 of the show's 156 episodes. He was also the show's host and narrator, delivering monologues at the beginning and end of each episode. Serling's opening and closing narrations usually summarize the episode's events encapsulating how and why the main character(s) had entered the Twilight Zone.

In 1997, the episodes "To Serve Man" (directed by Richard L. Bare) and "It's a Good Life" (directed by James Sheldon) were respectively ranked at 11 and 31 on TV Guide's 100 Greatest Episodes of All Time. Serling himself stated that his favorite episodes of the series were "The Invaders" (directed by Douglas Heyes) and "Time Enough at Last" (directed by John Brahm).

The Twilight Zone is widely regarded as one of the greatest television series of all time. In 2002, the series was ranked No. 26 on TV Guide's 50 Greatest TV Shows of All Time. In 2004, it was ranked No. 8 on TV Guide's Top Cult Shows Ever, moving to No. 9 three years later. In 2013, the Writers Guild of America ranked it as the third best-written TV series ever and TV Guide ranked it as the fourth greatest drama, the second greatest sci-fi show and the fifth greatest show of all time. In 2016, the series was ranked No. 7 on Rolling Stone's list of the 100 greatest shows of all time and was ranked No. 12 in 2022.

Development

By the late 1950s, Rod Serling was a prominent name in American television. His successful television plays included Patterns (for Kraft Television Theatre) and Requiem for a Heavyweight (for Playhouse 90), but constant changes and edits made by the networks and sponsors frustrated Serling. In Requiem for a Heavyweight, the line "Got a match?" had to be struck because the sponsor sold lighters; other programs had similar striking of words that might remind viewers of competitors to the sponsor, including one case in which the sponsor, Ford Motor Company, had the Chrysler Building removed from a picture of the New York City skyline.

According to comments in his 1957 anthology Patterns, Serling had been trying to delve into material more controversial than his works of the early 1950s. This led to Noon on Doomsday for the United States Steel Hour in 1956, a commentary by Serling on the defensiveness and total lack of repentance he saw in the Mississippi town where the murder of Emmett Till took place. His original script closely paralleled the Till case, then was moved out of the South and the victim changed to a Jewish pawnbroker, and eventually watered down to just a foreigner in an unnamed town. Despite bad reviews, activists sent numerous letters and wires protesting the production.

Serling thought that a science-fictional setting, with robots, aliens and other supernatural occurrences, would give him more freedom and less interference in expressing controversial ideas than more realistic settings. "The Time Element" was Serling's 1957 pilot pitch for his show, a time travel adventure about a man who travels back to Honolulu in 1941 and unsuccessfully tries to warn everyone about the impending attack on Pearl Harbor. The script, however, was rejected and shelved for a year until Bert Granet discovered and produced it as an episode of Desilu Playhouse in 1958. The show was a great success and enabled Serling to finally begin production on his anthology series, The Twilight Zone. Serling's editorial sense of ironic fate in the writing done for the series was identified as significant to its success by the BFI Film Classics library which stated that for Serling "the cruel indifference and implacability of fate and the irony of poetic justice" were recurrent themes in his plots.

Episodes

Main article: List of The Twilight Zone (1959 TV series) episodes

 

Season	Episodes		Originally aired
			First aired	Last aired
Concept			November 24, 1958
1	36		October 2, 1959	July 1, 1960
2	29		September 30, 1960	June 2, 1961
3	37		September 15, 1961	June 1, 1962
4	18		January 3, 1963	May 23, 1963
5	36		September 27, 1963	June 19, 1964

Season 1 (1959–60)

Main article: The Twilight Zone (1959 TV series, season 1)

There is a fifth dimension, beyond that which is known to man. It is a dimension as vast as space and as timeless as infinity. It is the middle ground between light and shadow, between science and superstition, and it lies between the pit of man's fears and the summit of his knowledge. This is the dimension of imagination. It is an area which we call The Twilight Zone.

— Rod Serling

Serling working on his script with a dictating machine, 1959

The Twilight Zone premiered the night of October 2, 1959, to rave reviews. "Twilight Zone is about the only show on the air that I actually look forward to seeing. It's the one series that I will let interfere with other plans", said Terry Turner for the Chicago Daily News. Others agreed. Daily Variety ranked it with "the best that has ever been accomplished in half-hour filmed television" and the New York Herald Tribune found the show to be "certainly the best and most original anthology series of the year".

Even as the show proved popular to television's critics, it struggled to find a receptive audience of television viewers. CBS was banking on a rating of at least 21 or 22, but its initial numbers were much worse. The series' future was jeopardized when its third episode, "Mr. Denton on Doomsday" earned a 16.3 rating. Still, the show attracted a large enough audience to survive a brief hiatus in November, after which it finally surpassed its competition on ABC and NBC and persuaded its sponsors (General Foods and Kimberly-Clark) to stay on until the end of the season.

With one exception ("The Chaser"), the first season featured scripts written only by Rod Serling, Charles Beaumont or Richard Matheson. These three were responsible for 127 of the 156 episodes in the series. Additionally, with one exception ("A World of His Own"), Serling never appeared on camera during any first-season episode (as he would in future seasons) and was present only as a voice-over narrator. Serling did appear on screen in Twilight Zone promotional spots plugging the following week's episode – just not in the episodes themselves. These promo spots were unseen for several decades after their initial airings; while many have been released in the DVD and Blu-ray releases of The Twilight Zone, a few are lost completely and some survive only as audio tracks. Most are available through Paramount+ when watching the full episodes.

Many of the season's episodes proved to be among the series' most celebrated, including "Time Enough at Last", "The Monsters Are Due on Maple Street", "Walking Distance", and "The After Hours". The first season won Serling an unprecedented fourth Emmy Award for dramatic writing, a Producers Guild Award for Serling's creative partner Buck Houghton, a Directors Guild Award for John Brahm and the Hugo Award for best dramatic presentation.

Bernard Herrmann's original opening theme music lasted throughout the first season. For the final five episodes of the season, the show's original surrealist "pit and summit" opening montage and narration was replaced by a piece featuring an eye that closed (revealing the setting sun) and shorter narration, and a truncated version of Herrmann's theme.

Some first-season episodes were available for decades only in a version with a pasted-on second-season opening. These "re-themed" episodes were prepared for airing in the summer of 1961 as summer repeats; the producers wanted to have a consistent opening for the show every week. During the original 1959/60 run, Herrmann's theme was used in every first-season episode. The first season openings for these episodes have since been restored to recent DVD and Blu-ray reissues although incorrect openings were restored on two episodes, "Mr. Denton on Doomsday" and "A Passage for Trumpet".

Season 2 (1960–61)

Main article: The Twilight Zone (1959 TV series, season 2)

You're traveling through another dimension, a dimension not only of sight and sound but of mind; a journey into a wondrous land whose boundaries are that of imagination. That's the signpost up ahead—your next stop, the Twilight Zone.

— Rod Serling

Serling models an airplane with actress Inger Stevens, who appeared in "The Hitch-Hiker" and "The Lateness of the Hour."

Pippa Scott in "The Trouble With Templeton"

The second season premiered on September 30, 1960, with "King Nine Will Not Return," Serling's fresh take on the pilot episode "Where Is Everybody?" The familiarity of this first story stood in stark contrast to the novelty of the show's new packaging: Bernard Herrmann's stately original theme was replaced by Marius Constant's more jarring and dissonant (and now more-familiar) new guitar-and-bongo theme. The closing eye was replaced by a more surreal introduction inspired by the new images in Serling's narration (such as "That's the signpost up ahead"), and Serling himself stepped in front of the cameras to present his opening narration, rather than being only a voice-over narrator (as in the first season). The openings of the first three episodes of the season retained the eye opening's narration.

A new sponsor, Colgate-Palmolive, replaced the previous year's Kimberly-Clark (as Liggett & Myers would succeed General Foods, in April 1961), and a new network executive, James Aubrey, took over CBS. "Jim Aubrey was a very, very difficult problem for the show," said associate producer Del Reisman. "He was particularly tough on The Twilight Zone because for its time it was a particularly costly half-hour show… Aubrey was real tough on [the show's budget] even when it was a small number of dollars." In a push to keep the show's expenses down, Aubrey ordered that seven fewer episodes be produced than last season and that six of those being produced would be shot on videotape rather than film, a move Serling disliked, calling it "neither fish nor fowl." Two additional episodes filmed in the second season ("The Grave" and "Nothing in the Dark") were held over to the third season.

Season two saw the production of many of the series' most acclaimed episodes, including "Eye of the Beholder," "Nick of Time," "The Invaders," "The Trouble With Templeton" and "Will the Real Martian Please Stand Up?." The trio of Serling, Matheson and Beaumont began to admit new writers, and this season saw the television debut of George Clayton Johnson. Emmys were won by Serling (his fifth) for dramatic writing and by director of photography George T. Clemens and, for the second year in a row, the series won the Hugo Award for best dramatic presentation. It also earned the Unity Award for "Outstanding Contributions to Better Race Relations" and an Emmy nomination for "Outstanding Program Achievement in the Field of Drama." The Twilight Zone was mentioned in Newton Minow's landmark 1961 speech "Television and the Public Interest" as one of the few quality television series on the air at the time in a "vast wasteland" of mass-produced junk, with Minow praising the series as "dramatic and moving."

Rod Serling at home in 1959

Five weeks into season two, the show's budget was showing a deficit. The total number of new episodes was projected at twenty-nine, more than half of which, sixteen, had already been filmed by November 1960. As a cost-cutting measure, six episodes ("The Lateness of The Hour," "The Night of The Meek," "The Whole Truth," "Twenty-Two," "Static," and "Long Distance Call") were produced in the cheaper videotape format, which also required fewer camera movements. In addition, videotape was a relatively primitive medium in the early 1960s; the editing of tape was next to impossible. Each of the episodes was, therefore "camera-cut" as in live TV—on a studio sound stage, using a total of four cameras. The requisite multi-camera setup of the videotape experiment made location shooting difficult, severely limiting the potential scope of the story-lines. Even with those artistic sacrifices, the eventual savings amounted to only $6,000 per episode, far less than the cost of a single episode. The experiment was not attempted again. Kinescope versions of the videotaped episodes were rerun in syndication.

Season 3 (1961–62)

Main article: The Twilight Zone (1959 TV series, season 3)

You're traveling through another dimension, a dimension not only of sight and sound but of mind; a journey into a wondrous land whose boundaries are that of imagination. Your next stop...the Twilight Zone.

— Rod Serling

In his third year as executive producer, host, narrator and primary writer for The Twilight Zone, Serling was beginning to feel exhausted. "I've never felt quite so drained of ideas as I do at this moment," said the 37-year-old playwright at the time. In the first two seasons he contributed 48 scripts, or 73% of the show's total output; he contributed 56% of this season's output. "The show now seems to be feeding off itself", said a Variety reviewer of the season's episode two. Sponsors for this season included Chesterfield, Bufferin tablets, and Pepsi-Cola.

Despite his avowed weariness, Serling again managed to produce several teleplays that are widely regarded as classics, including "It's a Good Life", "To Serve Man", "Little Girl Lost" and "Five Characters in Search of an Exit". Scripts by Montgomery Pittman and Earl Hamner, Jr. supplemented Matheson and Beaumont's output, and George Clayton Johnson submitted three teleplays that examined complex themes. The episode "I Sing the Body Electric" was contributed by sci-fi writer Ray Bradbury. By the end of the season, the series had reached over 100 episodes.

The Twilight Zone received two Emmy nominations (for cinematography and art design), but was awarded neither. It again received the Hugo Award for "Best Dramatic Presentation", making it the only three-time recipient until it was tied by Doctor Who in 2008.

In spring 1962, The Twilight Zone was late in finding a sponsor for its fourth season and was replaced on CBS's fall schedule with a new hour-long situation comedy called Fair Exchange. In the confusion that followed this apparent cancellation, producer Buck Houghton left the series for a position at Four Star Productions. Serling meanwhile accepted a teaching post at Antioch College, his alma mater. Though the series was eventually renewed, Serling's contribution as executive producer decreased in its final seasons.

Season 4 (1963)

Main article: The Twilight Zone (1959 TV series, season 4)

You unlock this door with the key of imagination. Beyond it is another dimension: a dimension of sound, a dimension of sight, a dimension of mind. You're moving into a land of both shadow and substance, of things and ideas; you've just crossed over into the Twilight Zone.

— Rod Serling

Julie Newmar and Albert Salmi in "Of Late I Think of Cliffordville."

In November 1962, CBS contracted Twilight Zone (now sans The) as a mid-season January replacement for Fair Exchange, the very show that replaced it in the September 1962 schedule. In order to fill the Fair Exchange time slot, each episode had to be expanded to an hour, an idea which did not sit well with Serling, nor the production crew. "Ours is the perfect half-hour show... If we went to an hour, we'd have to fleshen our stories, soap opera style. Viewers could watch fifteen minutes without knowing whether they were in a Twilight Zone or Desilu Playhouse," Serling responded. Herbert Hirschman was hired to replace long-time producer Buck Houghton. One of Hirschman's first decisions was to direct a new opening sequence, this one illustrating a door, eye, window and other objects suspended in space. His second task was to find and produce quality scripts. Sponsors included Johnson & Johnson.

This season of Twilight Zone once again turned to the reliable trio of Serling, Matheson and Beaumont. However, Serling's input was limited this season; he still provided the majority of the teleplays, but as executive producer, he was virtually absent and as host, his artful narrations had to be shot back-to-back against a gray background during his infrequent trips to Los Angeles. Due to complications from a developing brain disease, Beaumont's input also began to diminish significantly. Additional scripts were commissioned from Earl Hamner, Jr. and Reginald Rose to fill in the gap.

With five episodes left in the season, Hirschman received an offer to work on a new NBC series called Espionage and was replaced by Bert Granet, who had previously produced "The Time Element". Among Granet's first assignments was "On Thursday We Leave for Home," which Serling considered the season's most effective episode. There was an Emmy nomination for cinematography and a nomination for the Hugo Award.

Season 5 (1963–64)

Main article: The Twilight Zone (1959 TV series, season 5)

Serling later claimed, "I was writing so much, I felt I had begun to lose my perspective on what was good and what was bad". By the end of this final season, he had contributed 92 scripts in five years. This season, the new alternate sponsors were American Tobacco and Procter & Gamble. The show returned to its half-hour format.

Beaumont was now out of the picture almost entirely, contributing scripts only through the ghostwriters Jerry Sohl and John Tomerlin, and after producing only 13 episodes, Bert Granet left and was replaced by William Froug—with whom Serling had worked on Playhouse 90.

William Shatner in "Nightmare at 20,000 Feet."

Froug made a number of unpopular decisions; first by shelving several scripts purchased under Granet's term (including Matheson's "The Doll," which was nominated for a Writer's Guild Award when finally produced in 1986 on Amazing Stories); secondly, Froug alienated George Clayton Johnson when he hired Richard deRoy to completely rewrite Johnson's teleplay Tick of Time, eventually produced as "Ninety Years Without Slumbering." "It makes the plot trivial," complained Johnson of the resulting script, insisting he be given screen credit for the final version of the episode as "Johnson Smith." Tick of Time became Johnson's final submission to The Twilight Zone.

Even under these conditions, several episodes were produced that are well remembered, including "Nightmare at 20,000 Feet," "A Kind of a Stopwatch", "The Masks" and "Living Doll." Although this season received no Emmy recognition, episode number 142, "An Occurrence at Owl Creek Bridge"—a 1962 French-produced short film which was modified slightly for broadcast—received the Academy Award for best short film in 1963.

In late January 1964, CBS announced the show's cancellation. "For one reason or other, Jim Aubrey decided he was sick of the show… [H]e claimed that it was too far over budget and that the ratings weren't good enough", explained Froug. But Serling countered by telling the Daily Variety that he had "decided to cancel the network". ABC showed interest in bringing Serling over to their network to write a more explicitly horror-themed series, Witches, Warlocks and Werewolves, but Serling was not impressed. "The network executives seem to prefer weekly ghouls, and we have what appears to be a considerable difference in opinion. I don't mind my show being supernatural, but I don't want to be booked into a graveyard every week." Shortly afterwards, Serling sold his 40% share in The Twilight Zone to CBS, leaving the show and all projects involving the supernatural behind him until 1969, when Night Gallery debuted.

Casting

Main article: List of The Twilight Zone (1959 TV series) guest stars

Being an anthology series with no recurring characters, The Twilight Zone features a wide array of guest stars for each episode, some of whom appeared in multiple episodes. Many episodes feature early performances from actors who later became famous, such as Buddy Ebsen, Telly Savalas, Theodore Bikel, Bill Bixby, Lloyd Bochner, Morgan Brittany, Charles Bronson, Carol Burnett, Donna Douglas, Robert Duvall, Peter Falk, Constance Ford, Joan Hackett, Dennis Hopper, Ron Howard, Jim Hutton, Jack Klugman, Martin Landau, Cloris Leachman, Jean Marsh, Elizabeth Montgomery, Billy Mumy, Julie Newmar, Barbara Nichols, Leonard Nimoy, Robert Redford, Burt Reynolds, Janice Rule, William Shatner, Dean Stockwell, George Takei, Joyce Van Patten, Jack Warden, Jonathan Winters, and Dick York. Other episodes feature performances by actors later in their careers, such as Dana Andrews, Joan Blondell, Ann Blyth, Art Carney, Jack Carson, Gladys Cooper, William Demarest, Andy Devine, Cedric Hardwicke, Josephine Hutchinson, Buster Keaton, Ida Lupino, Kevin McCarthy, Burgess Meredith, Agnes Moorehead, Alan Napier, Franchot Tone, Mickey Rooney, and Ed Wynn. Klugman and Meredith are tied for the most starring roles with a record of four episodes.

Character actors who appeared (some more than once) include John Anderson, John Dehner, Betty Garde, Sandra Gould, Nancy Kulp, Celia Lovsky, Eve McVeagh, Nehemiah Persoff, Albert Salmi, Vito Scotti, Olan Soule, Harold J. Stone, and Estelle Winwood. The actor who appears in the most episodes is Robert McCord.

Music

Besides Bernard Herrmann and Jerry Goldsmith, other contributors to the music were Nathan Van Cleave, Leonard Rosenman, Fred Steiner, and Franz Waxman. The first season featured an orchestral title theme by Herrmann, who also wrote original scores for seven of the episodes, including the premiere, "Where Is Everybody?". The guitar theme most associated with the show was written by the French avant-garde composer Marius Constant as part of a series of short cues commissioned by CBS as "work made for hire" library music for the series. The guitar player was Howard Roberts. Used from season two onward, the theme as aired was a splicing together of two of these library cues: "Etrange 3 (Strange No. 3)" and "Milieu 2 (Middle No. 2)". Varèse Sarabande released several albums of music from the series, focusing on the episodes that received original scores.

Volume 1

1. Main Title Theme – Marius Constant (:27)
2. The Invaders – Jerry Goldsmith (12:57)
3. Perchance To Dream – Nathan Van Cleave (9:52)
4. Walking Distance – Bernard Herrmann (12:52)
5. The Sixteen-Millimeter Shrine – Franz Waxman (10:55)
6. End Title Theme – Marius Constant (:42)

Volume 2

1. Main Title Theme – Bernard Herrmann (1:11)
2. Where Is Everybody? – Bernard Herrmann (11:19)
3. 100 Yards Over The Rim – Fred Steiner (12:14)
4. The Big Tall Wish – Jerry Goldsmith (11:52)
5. A Stop at Willoughby – Nathan Scott (12:24)
6. End Title Theme – Bernard Herrmann (1:05)

Volume 3

1. Alternate Main Title Theme – Marius Constant (:38)
2. Back There – Jerry Goldsmith (12:51)
3. And When The Sky Was Opened – Leonard Rosenman (11:54)
4. A World Of Difference – Nathan Van Cleave (11:48)
5. The Lonely – Bernard Herrmann (11:09)
6. Alternate End Title – Marius Constant (:54)

Volume 4

1. Alternate Main Title – Bernard Herrmann (:28)
2. Jazz Theme One – Jerry Goldsmith (9:12)
3. Jazz Theme Two – Jerry Goldsmith (3:12)
4. Jazz Theme Three – Rene Garriguenc (4:04)
5. Nervous Man in a Four Dollar Room – Jerry Goldsmith (8:16)
6. Elegy – Nathan Van Cleave (8:14)
7. King Nine Will Not Return – Fred Steiner (11:11)
8. Two – Nathan Van Cleave (12:09)
9. Alternate End Title – Bernard Herrmann (:43)

Volume 5

1. Alternate Main Title #2 – Bernard Herrmann (:29)
2. I Sing The Body Electric – Nathan Van Cleave (11:41)
3. The Passerby – Fred Steiner (12:58)
4. The Trouble With Templeton – Jeff Alexander (11:46)
5. Dust – Jerry Goldsmith (11:33)
6. Alternate End Title #2 – Bernard Herrmann (1:07)

Many of the above were included on a four-disc set released by Silva America. Varese also released a two-disc set of re-recordings of Herrmann's seven scores for the series ("Where Is Everybody?", "Walking Distance", "The Lonely", "Eye of the Beholder", "Little Girl Lost", "Living Doll", and "Ninety Years Without Slumbering"), conducted by Joel McNeely. Alongside this release, Bernard Herrmann's score for the episode "Walking Distance" received another re-recording accompanying a new recording of his score for François Truffaut's "Fahrenheit 451" performed by the Moscow Symphony Orchestra, conducted by William T. Stromberg and released by Tribute Film Classics.

1961 LP record release

In that year, Marty Manning And His Orchestra released an LP record The Twilight Zone: A Sound Adventure In Space on Columbia Records. It was recorded with top New York City session musicians, including Mundell Lowe (guitar), Jerry Murad (harmonica), Harry Breuer (vibraphone), and Phil Kraus (percussion). Lyric soprano Lois Hunt provided the wordless vocals, and Teo Macero was credited with special effects. Manning himself was credited with playing the serpent, Ondioline, and Ondes Martenot.

The first track was the title theme. Thereafter, the other tracks, and their writers, were:-

side A

1. The Twilight Zone (2:07)

   Written-By – M. Manning

2. Forbidden Planet (2:28)

   Written-By – D. Rose

3. The Lost Weekend Theme (2:41)

   Written-By – Miklos Rozsa

4. Invitation (3:04)

   Written-By – B. Kaper

5. You Stepped Out Of A Dream (2:16)

   Written By – Gus Kahn-N.H. Brown

6. The Unknown (2:15)

   Written-By – M. Manning

side B

1. Far Away Places (2:13)

   Written By – J. Whitney-A.C. Kramer

2. Spellbound Concerto (2:16)

   Written-By – Miklos Rozsa

3. The Sorcerer's Apprentice (2:16)

   Arranged By – Marty Manning

   Composed By – Dukas

4. The Moon Is Low (2:25)

   Written By – A. Freed-N.H. Brown

5. Night On Bald Mountain (2:19)

   Arranged By – Marty Manning

   Composed By – Mussorgsky

6. Shangri-La (n/a)

   Written By – R. Maxwell-M. Malneck

Broadcast history

Season	Time slot
1 (1959–1960)	Friday at 10:00-10:30 pm E.T.
2 (1960–1961)
3 (1961–1962)
4 (1963)	Thursday at 9:00-10:00 pm E.T.
5 (1963–1964)	Friday at 9:30-10:00 pm E.T.

Awards and nominations

The Twilight Zone was nominated for four Primetime Emmy Awards, winning two.

Year	Association	Category	Nominee	Result
1960	Primetime Emmy Awards	Outstanding Writing Achievement in Drama	Rod Serling	Won
1961		Outstanding Program Achievement in the Field of Drama	The Twilight Zone	Nominated
		Outstanding Writing Achievement in Drama	Rod Serling	Won
1962		Outstanding Writing Achievement in Drama	Rod Serling	Nominated
1963	Golden Globe Awards	Best TV Producer/Director	Rod Serling	Won

In media

Syndication

Most episodes continue to be broadcast in syndication. After the cancellation of the series, Serling sold his rights to CBS, unaware of what the future would hold in syndication, and the royalties he would have gained.

Episodes are broadcast nationally on the Syfy channel in the United States. They are regularly shown in late-night slots and in marathons aired typically every year on New Year's Eve and Day and the Fourth of July. Syfy broadcasts are often re-cut to feature more commercials during the time slot, in order to meet the 22 or 44-minute maximum episode runtime.

Originally, there were five episodes not included in the syndication package. Three of those ("Sounds and Silences", "Miniature", and "A Short Drink From a Certain Fountain") were involved in copyright infringement lawsuits. The other two, which have never been in syndication (both from season five), are "An Occurrence at Owl Creek Bridge" (a French short film, aired twice per agreement with the filmmakers) and "The Encounter" (which was pulled after its initial showing, due to the racial overtones). "The Encounter" has since aired on Syfy for the first time in 2016.

Home media

The Twilight Zone was released on Region 1 DVD for the first time by Image Entertainment. All of the releases feature uncut episodes. The season releases (The Definitive Collection and Blu-rays) also include the radio dramas and the "Next Week" promos (some of the promos on the season DVDs are audio only). The various releases include:

• 43 volumes of 3 to 4 episodes each (released December 29, 1998 – June 12, 2001)
• Five 9-disc Collection DVD sets (released December 3, 2002 – February 25, 2003)
• Season sets: The Twilight Zone: The Definitive Collection (released December 28, 2004 – December 26, 2005)
• The Twilight Zone: The Complete Definitive Collection, 28 discs (released October 3, 2006)
• The Twilight Zone: The Complete Series (Episodes Only Collection), 25 discs (released November 19, 2013; reissued November 11, 2016)

Compilations

• Treasures of The Twilight Zone (3-episode compilation released November 24, 1997)
• More Treasures of The Twilight Zone (3-episode compilation released November 24, 1998)
• The Twilight Zone: 40th Anniversary Gift Pack (19-episode compilation released September 21, 1999)
• The Twilight Zone: Fan Favorites (19-episode compilation released October 26, 2010)
• The Twilight Zone: More Fan Favorites (20-episode compilation released May 8, 2012)
• The Twilight Zone: Essential Episodes (17-episode compilation released July 4, 2014; reissued October 11, 2016)

Limited set

• The Twilight Zone: Gold Collection, a 49-disc set of the entire series, released by V3 Media on December 2, 2002 – only 2,500 copies of this set were made.

Blu-ray
Note: all of the Blu-ray releases are Region A

• The Twilight Zone: Season 1 (released September 14, 2010)
• The Twilight Zone: Season 2 (released November 16, 2010)
• The Twilight Zone: Season 3 (released on February 15, 2011)
• The Twilight Zone: Season 4 (released on May 17, 2011)
• The Twilight Zone: Season 5 (released on August 30, 2011)
• The Twilight Zone: The Complete Series, 24 discs (released on June 5, 2012; reissued December 13, 2016)

The 1958 Desilu Playhouse episode, "The Time Element," considered to be a "first" pilot for The Twilight Zone (see above) is included as a bonus feature on the Blu-ray release (with Season 1), but not on any of the earlier DVD releases.

UK release

Fremantle Media released a box set for each season of The Twilight Zone on both DVD and Blu-ray over 2011 and early 2012. These sets received high praise and won an award from The Guardian for Best Special Features of 2011. These Blu-rays and DVDs are multi-region and so can be played around the world.

Radio

Main article: The Twilight Zone (radio series)

In 2002, the BBC engaged producer Carl Amari to license the rights from the Rod Serling Estate to turn the TV series into a weekly radio drama series for BBC Radio 4 Extra which in turn was purchased and distributed by CBS Enterprises in the US. The series features Stacy Keach in Rod Serling's role as narrator and each 40-minute audio drama includes a Hollywood celebrity in the starring role. Some of the stars include Jim Caviezel, Blair Underwood, Jason Alexander, Jane Seymour, Lou Diamond Phillips, Luke Perry, Michael York, Sean Astin, and Ernie Hudson. The episodes air nationally on hundreds of radio stations and Sirius/XM, and are available for download.

Online distribution

As of April 2019, all half-hour episodes (seasons 1–3 and 5) of the series are available on Netflix Instant Streaming in Brazil, Mexico, and the U.S.

All five seasons of the series are available on Hulu, Amazon Video, and iTunes.

All seasons as aired, including promotional spots recorded by Mr. Serling, are available on Paramount+.

Revivals

The series has seen three revivals:

• The Twilight Zone (1985 TV series)
• The Twilight Zone (2002 TV series)
• The Twilight Zone (2019 TV series)