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Wednesday, January 29, 2014

Extending Fermat to Other Powers

File:Pierre de Fermat.jpg

January 29'th, David Strumfels, A Medley of Potpourri blog

We all know Fermat's famous Last Theorem, solved almost 20 years ago (!) by Andrew Wiles (using mathematical techniques the Fermat did not have, so his proof remains a mystery to history).  The theorem states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two.  Examples:  3^2 + 4^2 = 5^2, and 12^2 + 5^2 = 13^2.

I'd been toying around with variations of combining various integers raised to various powers for some time (when you enjoy doing something you don't notice how much time), when I observed that:
3^3 + 4^3 + 5^3 = 6^3.  In other words, here I had found (undoubtedly not for the first time in history) a group of four positive integers, a, b, and c, can satisfy the equation a^n + b^n + c^n = d^n whenever n = three.

But does Fermat's modified Theorem apply here too?  Can n never exceed three?  Could a similar method be used to prove this Theorem?  And furthermore, does the fact of squares and cubes having these relationships, mean they keep on going up the line, infinitely.  E.g., can five integers, raised to the fourth power, be found with this relationship?  And on and on?  (The 5/4 set is false for 2,3,4,5,6, if you are curious; try it.)

Now I am no mathematician, but a little math instinct tells me something interesting is going on here.  A very large theorem, encompassing Fermat's and our third power analogue and possibly beyond feels ... well, like a genuine mathematical conjecture at least, if I use the word correctly.  First, I shall look for a fourth power analogue, for if it doesn't exist then I am blowing smoke (I have to assume it will be found, if at all, with fairly small integers, as with the second and third powers.)

I will work on this, and feel free to give it your all too, if you want to.  That's all for now.

David J. Strumfels

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