= 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. Consequently
$$
\boxed{z^{2n}\equiv\begin{cases}0\pmod3,&3\mid z,\\1\pmod3,&3\nmid z.\end{cases}}
$$
The exponent in the original PDF is $2n$; the polynomial identities in the converted TeX are a transcription error. The argument uses $n\ge1$; it does not assign $0^0$ a value.
Back to article page
