Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 1D (Numbers and Sets)

Reduce $z$ using modular congruence: its residue class modulo $3$ is $0$, $1$, or $-1$. A square number therefore has residue class $0$ or $1$, and raising either residue to the positive integer power $n$ leaves it unchanged. ConsequentlyThe exponent in the original PDF is $2n$; the polynomial identities in the converted TeX are a transcription error. The argument uses $n\ge 1$; it does not assign $0^0$ a value.

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 $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus 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 $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.