Using Wilson's theorem show if p is prime and $p\equiv 1(mod4)$ then $x^2\equiv -1(modp)$ has 2 incongruent solutions

Question

Using Wilson's theorem show if $p$ is prime and $p\equiv 1\pmod4$ then $x^2\equiv -1\bmod p$ has 2 incongruent solutions $x\equiv \pm((p-1)/2)!\pmod p$

No idea how to do this