Diophantine Quadratic Equation in Three Variables

Prove that for all positive integers n, the equation

x² + y² + z² = 59ⁿ

is solvable in positive integers.

Prove that for all positive integers n, the equation

x² + y² + z² = 59ⁿ

is solvable in positive integers.

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 a 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.

References

T. Andreescu, D. Andrica, I. Cucurezeanu, An Introduction to Diophantine Equations, Birkhäuser, 2010, p. 38

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