# Fermat's Like Equaition

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

For which positive integers k does the equation

X^{2k-1} + y^{2k} = z^{2k+l}

have a solution in positive integers x, y, and z?

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^{m}2^{2k} = a^{m}3^{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.

|Contact| |Front page| |Contents| |Algebra|

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

69118843