 Solution

True. Write the odd integer as $m=2n+1$. The identity from part (b) and $p\mid F_m$ giveConsecutive Fibonacci numbers are coprime, so $F_{n+1}$ is invertible modulo $p$. HenceThus $-1$ is a quadratic residue modulo the odd prime $p$, which occurs exactly when $\boxed{p\equiv 1(\mathrm{mod}4)}$.