 Solution

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.