Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 5E / c

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 $\boxed{p\equiv 1(\mathrm{mod}4)}$.

This statement can be false. For example, $m=10$ is even andbut the odd prime divisor $p=5$ satisfies $p\equiv 1(\mathrm{mod}4)$, rather than $3(\mathrm{mod}4)$.