Solution
= Solution
\b[True.] Write the odd integer as $m=2n+1$. The identity from part (b) and $p\mid F_m$ give
$$
F_n^2+F_{n+1}^2\equiv0\pmod p.
$$
Consecutive <Fibonacci numbers> are coprime, so $F_{n+1}$ is invertible modulo $p$. Hence
$$
(F_nF_{n+1}^{-1})^2\equiv-1\pmod p.
$$
Thus $-1$ is a <quadratic residue> modulo the odd prime $p$, which occurs exactly when $\boxed{p\equiv1\pmod4}$.