The problem goes as :
Let $p$ be an odd prime & $\mathbb F_{p} =\mathbb Z/p\mathbb Z$.Show that: $x^{2}+1$ has a root in $\mathbb F_{p}$ iff $p \equiv 1 ( mod $ $4)$ .
My Solution: $\mathbb F^{*}_{p} $is the corresponding multiplicative group of order $p-1$ . Now $x^{2}+1$ has a root $t$ in $\mathbb F_{p}$ iff $\exists $ $t$ $\in \mathbb F_{p}$ satisfying: $t^{2}=-1$ . Clearly then, $t \in F^{*}_{p}$ . i.e. $ o(t) | p-1$. Now, in $F^{*}_{p}$ , the identity element is: $1 $ . & obviously $t^{4} =1$ which implies $o(t) = 4$ . i.e. $4|p-1$ . Thus, $x^{2}+1$ has a root in $\mathbb F_{p}$ iff $p \equiv 1 ( mod $ $4)$.
[Q.E.D]
Is my solution is complete?? & Is there anything to be done to take care of the "iff" mentioned in question??
You have proved that if the root exists, then $p \equiv 1 \pmod{4}$. What remains is to show that for every such $p$, the root in $\mathbb{F}_p$ exists.
Hint: To prove this other implication, you can use Wilson's theorem.