# Let n∈ℕ. Suppose that p is an odd prime number that divides n^2+1 Show that if p=4k+3 for some integer k≥0, then n^(p−1)≡−1(modp).

Let $n \in \mathbb N$. Suppose that $p$ is an odd prime number that divides $n^2+1$. Show that if $p=4k+3$ for some integer $k \ge 0$, then $n^{p−1}\equiv−1 \pmod p$.

The question says to use "if $p=4k+3$ then $\frac {(p-1)} 2$ is odd".

So we know $n^2 \equiv -1 \pmod p$

Do I manipulate that into $n^{4k+2} \equiv -1 \pmod p$ ? I tried that and ended up with $n^{4k+2} \equiv -1^k \pmod p$.

How do I use "if $p=4k+3$ then $\frac {p-1} 2$ is odd" ?

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Modulo $p$ we get $n^{p-1} \equiv (n^2)^{(p-1)/2} \equiv -1^{(p-1)/2} \equiv -1$, which gives the implication you are looking for. By the way this implication is vacuously true, as the odd prime divisors of $n^2 + 1$ are all of the form $4k + 1$.