Show that if aⁿ+bⁿ= 2^m, and a, b, m, and n are positive integers (n>1), then a=b.

(In reply to Begin by Gamer)

Suppose a and b have a common prime factor p not 2. Then the left side is a multiple of k but the right side is clearly not. So if this equation is true, a and b cannot share a prime factor except perhaps 2.

Suppose a and b have different numbers of 2's in their factorization. Then we can factor out the smaller number of 2's to give a'^n + b'^n = 2^m' where m' = m - (smaller number of 2's). Now since a and b were assumed to have different numbers of 2's in their respective factorization, after dividing by the greatest common power of 2, the result has one odd and one even among a and b (if both were even, then there's another common 2 to factor out; if both were odd then the two numbers started with the same number of 2's) but the sum of an even and an odd is odd, so that makes the left side odd and the right side even. Another contradiction, so if the equation is true, then a and b have the same number of 2's in their factorizations AND (from above) have no other common factors.

So that means we have to consider a'^n + b'^n = 2^m' where a' and b' are both odd and are relatively prime to find solutions to the original equation. Now, n can't be odd, because if it were then (a'+b') would be a factor of the left side as would a polynomial of degree n-1 with n terms. Each of those n terms would be odd, being the product of nothing but a' and b' raised to various powers. So there would be an odd number of odd terms summed, which is always odd. Thus, if n is odd, then the left side has an odd factor, so n must be even.

If n is even, then a'^n and b'^n are both = 1mod8 and their sum = 2mod8. the only way this equation can hold is if a'=b'=1, as 2 is the only power of 2 that = 2mod8. And if a'=b'=1 then the original a and b have no common factors beyond their common factors of 2 which are equal. So, not only does a=b but a=b=2^z where z=(m-1)/n

This seems like a very convoluted approach. I'd love to see something simple and elegant that accomplishes the same goal!

Posted by Paul on 2007-04-07 03:40:34