Using Wilson's Theorem to find solution to $n^2 \equiv -1 \pmod{p}$

Question

I am currently studying number theory and its basic topics, and I came across this problem.

Wilson's Theorem states that if $p$ is prime, then $(p-1)!\equiv -1 \pmod{p}$.

If $p \equiv 1 \pmod{4}$ is prime, then use Wilson's Theorem to find a number $n$ so that $n^2 \equiv -1 \pmod{p}$.

I tried to substitute $p$ with $4k+1$, but I didn't get much far. I don't have any other ideas for tackling this problem. Any, hopefully elementary, solutions?

Note this is very similar, basically a duplicate, of If $p$ is prime and $p$ $\equiv$ $1$ (mod 4), then the congruence $x^2$ $\equiv$ $-1$ (mod $p$) has two incongruent solutions..., that contains a solution which uses the same approach as J. W. Tanner's answer below. — John Omielan, Jul 19 '20 at 21:14

score 3 · Accepted Answer · answered Jul 19 '20 at 21:11

By Wilson's theorem,

$(p-1)!=(p-1)\times(p-2)\times\cdots\times\dfrac{p+1}2\times\dfrac{p-1}2\times\cdots\times2\times1\equiv-1\pmod p.$

This is $(-1)(-2)\cdots\times(-\dfrac{p-1}2)\times\dfrac{p-1}2\times\cdots\times2\times1\equiv-1\pmod p,$

and, since it is essentially given that $\dfrac{p-1}2$ is even,

we can factor out $(-1)^{\text{even}}=1$ to get $\left(1\times2\times\cdots\times\dfrac{p-1}2\right)^2\equiv-1\pmod p.$

Using Wilson's Theorem to find solution to $n^2 \equiv -1 \pmod{p}$

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