[Anonymous314]

We will prove that $2|3^{2k} + 1$ and $4\nmid 3^{2k} + 1$.
By checking modulo we have
$3^{2k} + 1 \equiv 9^k + 1 \equiv (9 - 2 \cdot 4)^k + 1 \equiv 1 + 1 \equiv 2(\bmod 4)$.
$\therefore 2|3^{2k} + 1$ and $4\nmid 3^{2k} + 1$