Fermat's Like Equaition

Here's a problem proposed by Albert F. S. Wong (Mathematics Magazine, Vol. 77, No. 3 (Jun., 2004)):

The solution is by Jerry W. Grossman (Mathematics Magazine, Vol. 78, No. 3, (Jun., 2005)).

There are solutions for all k. Because 2k - 1, 2k, and 2k + 1 are pairwise relatively prime, it follows from the Chinese Remainder Theorem that there is a positive integer m with

m ≡ 0 (mod 2k),
m ≡ 0 (mod 2k + 1),
m ≡ -1(mod 2k - 1).

There are then positive integers r, s, t with

m = r(2k) = s(2k + 1) = t(2k - 1) - 1.

Now let a = 3^2k+1 - 2^2k, so a + 2^2k = 3^2k+1. Multiply through by a^m to obtain

a^m+1 + a^m2^2k = a^m3^2k+1.

This can be put into the form

(a^t)^2k-1 + (2a^r)^2k = (3a^s)^2k+1,

giving a solution to the diophantine equation.

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Copyright © 1996-2012 Alexander Bogomolny