TITLE: Elementary proof that $-1$ is a square in $\mathbb{F}_p$ for $p = 1 \mod{4}$
QUESTION [5 upvotes]: I am trying to proof that  $-1$ is a square in $\mathbb{F}_p$ for $p = 1 \mod{4}$. Of course, this is really easy if one uses the Legendre Symbol and Euler's criterion. However, I do not want to use those. In fact, I want to prove this using as little assumption as possible.
What I tried so far is not really helpful:
We can easily show that $\mathbb{F}_p^*/(\mathbb{F}_p^*)^2 = \{ 1\cdot (\mathbb{F}_p^*)^2, a\cdot (\mathbb{F}_p^*)^2 \}$ where $a$ is not a square (this $a$ exists because the map $x \mapsto x^2$ is not surjective). Now $-1 = 4\cdot k = 2^2 \cdot k$ for some $k\in \mathbb{F}_p$.
From here I am trying to find some relation between $p =1 \mod{4}$ and $-1$ not being a multiple of a square and a non-square.

REPLY [7 votes]: On $\mathbb{F}_p^\ast$, define the equivalence relation
$$x\sim y :\!\!\iff (x = y) \lor (x = -y) \lor (xy = 1) \lor (xy = -1).$$
Generally, the equivalence class of $x$ has four elements, $\{ x, -x, x^{-1}, -x^{-1}\}$. Since we always have $x \neq -x$, every class has at least two elements, and if $x = x^{-1}$ or $x = -x^{-1}$, then the equivalence class of $x$ has precisely two elements. The case $x = x^{-1}$ (or $x^2 = 1$) yields the class $\{1,-1\}$, so there is always at least one class with two elements.
If $p \equiv 1 \pmod{4}$, the number of elements of $\mathbb{F}_p^\ast$ is divisible by $4$, therefore the number of two-element classes must be even, hence $2$, and that means there must be an $x$ with $x = -x^{-1}$, or $x^2 = -1$.