Quick, self-contained way to see why $\left({{-1}\over p}\right) = 1$? Let $p$ be a prime number congruent to $1$ modulo $4$. What is a quick and self-contained way to see why$$\left({{-1}\over p}\right) = 1?$$
 A: On $(\mathbb Z/p\mathbb Z)^*=\{1,2, \dotsc, p-1\}$ consider the equivalence relation, defined by $x \sim y$ iff $x=-y$ or $x=y^{-1}$ or $x=-y^{-1}$.
Since $p$ is odd, we have $x \neq -x$, i.e. any equivalence class has $2$ or $4$ elements:


*

*If $x^2=1$, we have $[x] = \{x,-x\}$

*If $x^2=-1$, we have $[x] = \{x,-x=x^{-1}\}$

*In any other case, we have $[x] = \{x,-x,x^{-1},(-x)^{-1}\}$


We certainly have the equivalence class $\{1,p-1=-1\}$.
$p-1 \equiv 0 \pmod 4$ yields, that there will be another equivalence class with $2$ elements. This equivalence class must come from the second case, since the first case occurs only once ($x^2=1$ implies $x=\pm 1$ over a field). This shows that $x^2=-1$ is solvable.
A: If $p$ is any odd prime, then by Wilson's Theorem: $$(p-1)!\equiv 1\cdot 2\cdots\left(\frac{p-1}{2}\right)\left(-\frac{p-1}{2}\right)\cdots (-2)(-1)\pmod{p}$$
$$\equiv (-1)^{\frac{p-1}{2}}\left(\left(\frac{p-1}{2}\right)!\right)^2\equiv -1\pmod{p}$$
If $p\equiv 1\pmod{4}$, then all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm\left(\frac{p-1}{2}\right)!\pmod{p}$.

Edit: another proof: Let $p$ be prime, $p\equiv 1\pmod{4}$ and let $g$ be a primitive root mod $p$. Then $\text{ord}_p(g)=p-1$, so $g^{\frac{p-1}{2}}\equiv -1\pmod{p}$. Therefore, all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm g^{\frac{p-1}{4}}\pmod{p}$.
A: Suppose 
$$\;p=1\pmod4\implies p-1=0\pmod4\implies \;\text{the cyclic group}\;\;\Bbb F_p^*\;$$
 of order $\;p-1\;$ has a unique subgroup of order $\;4\;$, say $\;T:=\langle x\rangle\;$ , and since
$$\text{ord}\,(x)=4\implies x^2=-1$$
