If $4 | x^2+y^2+z^2$ then $x, y, z$ are all even

Question

I need to show that if $4 | x^2+y^2+z^2$ then $x, y, z$ are all even. I have absolutely no idea how to solve this question.

The second question I have is:

If $k,s \in \mathbb N$ then the equation $x^2+y^2+z^2 = 4^{s-1}(8k-1)$ does not have any integer solution.

My attempt: From the first question I can say that if $(x, y, z)$ is a solution then $x, y, z$ are even (because $s>1$, I can prove that there is no solution for $s=1$). but I don't know how to continue. I need any kind of help you can give.

Thank in advance

Answer 1

We know that each square is of the form $4n$ or $4n+1$. (The proof to this will be in a comment.) Use contradiction.

Say that there is at least one odd square, in the form $4k+1$. Then, no matter whether the second and third numbers are even or odd, the sum of squares will not be divisible by $4$. If we already assign a $4k+1$, then we need to add $3$ to get it to be divisible by $4$. But of course, no pair of $0$s or $1$s (the constant terms from $4n+\boxed{0}$ and $4n+\boxed{1}$) will add to $3$.

Therefore, all three squares must be even, and using this, I think you can solve the second problem.

Hint for second problem: If $s \ne 1$, then $x^2+y^2+z^2$ must be divisible by $4$, which we proved is impossible. Now you just have to prove for $s = 1$.

Answer 2

If $n$ is even then $n^2\equiv 0\pmod{4}$ and if $n$ is odd then $n^2\equiv 1\pmod{4}$.
This means that $x^2+y^2+z^2\pmod{4}\in\{0,1,2,3\}$ is exactly the number of odd numbers in $\{x,y,z\}$.