Show that $c^2 \equiv -1 \pmod p$ Let $p \equiv 1 \pmod4$ be a prime. Write $p$ in the form $p=a^2+b^2$ where $a$ and $b$ are integers. Let $c \equiv ab^{-1} \pmod p$. Show that $c^2 \equiv -1 \pmod p$
$p = 4n+1$ where $n$ is an integer.
$c = ab^{-1} +pm$ where $m$ is an integer.
Could someone give me a hint?

Since $c\equiv ab^{-1}$, we have $c^2\equiv a^2b^{-2}$.  Also:
$$a^2+b^2=p \implies a^2 + b^2 \equiv 0 \pmod p$$
Now multiply by $b^{-2}$ to get:
$$a^2b^{-2} + 1 \equiv 0\pmod p\implies c\equiv -1\pmod p$$
