= Solution
For a <Pythagorean triple>, each <square number> is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, <modular congruence> would give $z^2\equiv1+1\equiv2\pmod3$, impossible for a <square number>. Thus \b[at least one of $x,y$ is divisible by $3$]; both can be divisible, as in $(9,12,15)$.
An <odd number> has <square number> congruent to $1$ modulo $4$, whereas an <even number> has <square number> congruent to $0$. If both $x$ and $y$ were <odd numbers>, their sum of <square numbers> would again be $2$ modulo $4$, which no <square number> can be. Hence \b[$x$ and $y$ cannot both be odd]. They can both be even; the conclusion does not require the <Pythagorean triple> to be primitive.
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