Diophantine Quadratic Equation in Three Variables

Prove that for all positive integers n, the equation

x² + y² + z² = 59 ⁿ

is solvable in positive integers.

Solution

First of all observe that if (a, b, c) is a solution, with n = k, then (59a, 59b, 59c) is a solution for the same equation but with n = k + 2. Indeed,

(59a)² + (59b)² + (59c)² = 59²(a² + b² + c²) = 59²59^k = 59^{k + 2}.

From here, if we have solution for n = 1 we can derive the solvability of the equation for all n odd. Similarly, having a solution for n = 2 would imply the solvability for all even n.

For n = 1, we easily find that (5, 5, 3) is a solution. Given a little more time, we could find the solution (1, 3, 7) used in the book. Anyway, the induction tells us that, for any odd n > 0, the equation has a solution.

Obviously, for n = 2, finding a solution is a more difficult task. The equation is

a² + b² + c² = 59² = 3481.

This is the place where I would allow to use a calculator. One can spend pretty much time on trying to get past this hurdle. This is why I do not like this problem very much. Here's one solution (14, 39, 42). This allows us to complete the induction.