Can integers of the form $4^m(8n+7)$ be expressed as a sum of three squares? Is there a proof by induction for this question?

Question

Prove that an integer of the form $4^m(8n+7)$ cannot be written as a sum of three squares.

Answer 1

Sure, induct on $m$. Suppose $4^m\left(8n+7\right) = x^2 + y^2 + z^2$. If $m = 0$, it should be fairly easy for you to verify that there are no solutions by looking at both sides modulo $8$ and listing all possible squares modulo $8$ (thinking about what the sum of any three of them could/couldn't be). Then assume $m \geq 1$. By looking at both sides modulo $4$ and since squares are either congruent to $0$ or $1$ modulo $4$, we conclude that all of $x^2$, $y^2$ and $z^2$ are congruent to $0$ modulo $4$, hence all of $x$, $y$ and $z$ are even, hence $4^{m-1}\left(8n+7\right) = \left(\frac{x}{2}\right)^2 + \left(\frac{y}{2}\right)^2 + \left(\frac{z}{2}\right)^2$ is an expression of $4^{m-1}\left(8n+7\right)$ as the sum of three integer squares. This gives you your induction step.