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

X2k-1 + y2k = z2k+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 = 32k+1 - 22k, so a + 22k = 32k+1. Multiply through by am to obtain

am+1 + am22k = am32k+1.

This can be put into the form

(at)2k-1 + (2ar)2k = (3as)2k+1,

giving a solution to the diophantine equation.

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  • Diophantine Equations
  • Finicky Diophantine Equations
  • Diophantine Quadratic Equation in Three Variables
  • An Equation in Reciprocals
  • A Short Equation in Reciprocals
  • Minus One But What a Difference
  • Two-Parameter Solutions to Three Almost Fermat Equations
  • Chinese Remainder Theorem
  • Step into the Elliptic Realm
  • Fermat's Like Equation
  • Sylvester's Problem
  • Sylvester's Problem, A Second Look
  • Negative Coconuts
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