If p is a prime number of the form 4n+1, show that congruence $x^2 \equiv - 1 \space (mod \space p) $ If $p$ is a prime number of the form $4n+1, n $ being a natural number, then show that congruence $x^2 \equiv - 1 \space (mod \space p) $ is solvable
I went through Lagrange's theorem, Fermat's theorem,Index of subgroup concepts. No idea came to me. 
I tried to apply Fermat's theorem(If $p$ is prime number and x is any integer, then $a^{p-1} \equiv 1 ( mod \space p ) $
$$x^{4n+1-1} \equiv 1 (mod \space p)$$
$$x^{4n} \equiv 1 (mod \space p)$$
$$x^{4n} - 2 \equiv -1 (mod \space p)$$
I couldn't proceed from here onwards
I went through the answers in 
Show that if $p$ is a prime of the form $p=4n+1$, then we can solve $x^2\equiv -1\mod p$(with $x$ an integer).
i couldn't understand because I am not aware of Wilson's theorem. 
This problem is give as part of my group theory course. If possible, please provide answers from group theory point of view.
 A: Let $p$ be an odd prime. Let $P=\{1,...,p-1\}.$ Let $S$ be the set of squares  members  of $P$ that are congruent mod $p$ to squares of members of $P.$ 
For $x\in P$ let $f(x)\in S$ such that $x^2\equiv f(x)\pmod p.$ Then $f$ is exactly $2$-to-$1$  because if $x,x'\in P$ then $$f(x)=f(x')\iff p|(x-x')(x+x')\iff (x=x'\lor x+x'=p)\iff$$ $$\iff (x=x'\lor x=p-x'\ne x').$$ So $S$ has exactly $(p-1)/2$ members. 
If $p=4n+1$ with $n\in \Bbb N$ then multiplication mod $p$ on $S$ is a group with $2n$ members. 
Theorem. If $G$ is a finite group with an even number of members and identity element $1$ then there exists $y\in G$ with $1=y^2\ne y.$
By this theorem, let $y\in S$ with $y^2\equiv 1 \not \equiv y\pmod p.$ Let $x\in S$ with $x^2\equiv y \pmod p.$ Then $$0\equiv y^2-1\equiv x^4-1\equiv (x^2-1)(x^2+1)\equiv (y-1)(x^2+1)\pmod p$$ $$\text {and }\quad (y-1)\not \equiv 0\pmod p.$$ Therefore $x^2+1\equiv 0\pmod p.$
Proof of Theorem: Draw the multiplication table of $G.$ Remove the columns headed by each $y$ and $y'$ if $yy'=1$ and $y\ne y'.$ This removes an even number of columns, (Why?) and does not remove the column headed by $1.$ There is at least one $other$ column because there is an even number of columns. This column must be headed by some $y\ne 1$ such that $y^2=1.$  
