$x^2\equiv -1\pmod{\!p}$ solvable $\!\iff p\equiv 1 \pmod{\! 4}\,$ by group theory

Question

I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well?

Answer 1

I would sketch the group theoretic fact as follows: Let $G$ be a finite abelian group. Then the following are equivalent:

(1) For every $d$, there are at most $d$ solutions to $g^d=1$.

(2) $G$ is cyclic.

(3) For $n$ dividing $|G|$, there are precisely $\phi(n)$ elements of order $n$.

The only place that I know of where hypothesis (1) shows up naturally is considering the multiplicative groups of finite fields. From this we get:

There is an element of order $d$ in $\mathbb{F}_{p^r}$ if and only if $d| p^r-1$.

I'm not sure there is any more to say.

Answer 2

Perhaps you refer to the proof below, which contradicts Lagrange's or Fermat's little Theorem.

$\qquad \rm p = 4n\!+\!3\,$ prime $\,\Rightarrow\, \rm\bmod p\!:\ \color{#c00}{{-}1\equiv x^2}\, \overset{(\ \ )^{\large 2n+1}}\Longrightarrow \color{#c00}{-1}\equiv (\color{#c00}{x^2})^{ 2n+1}\!\equiv x^{ p-1} \Rightarrow\!\Leftarrow$

A finite field of order $\rm\:4n+3\:$ has subgroup of squares of order $\rm\:2n+1.\,$ But one easily shows that in any group of odd order $\rm\, 2n+1\,$ the equation $\rm\; x^2 = a\;$ has the solution $\rm\: a^{n+1}\;$ (due to Lagrange in $1769$). Specializing $\rm\; a = -1\:$ we conclude that $\rm\; x^2 = -1\:$ has no roots. On the other hand, the equation always has a root in a finite field of order $\,\rm 4n+1\,$ by the case $\,\rm k=4\,$ of Frobenius's theorem (1907) - which says that if $\,\rm k\,$ divides the order of a finite group $\rm G$ then $\,\rm k\,$ also divides the number of $\,\rm k$'th roots of $\,1\,$ in $\,\rm G$.

Lagrange's root calculation can be viewed special case of this method, i.e. raise both sides of $\,x^2=a\,$ to power $\,\frac{1}2\equiv -n\pmod{\!2n\!+\!1}\,$ to get $\,x\equiv a^{-n}\equiv a^{n+1}.\ \ $

Answer 3

Probably the simplest way to state a corresponding group theory result is the generalization of Frobenius's theorem mentioned by Bill Dubuque. The generalization says that if $G$ is a finite group, then the number of solutions in $G$ to $x^n=1$ is a multiple of $\gcd(|G|,n)$.

In the case of the congruence $x^2\equiv -1\pmod{p}$, the solutions to this congruence are exactly the solutions to $x^4\equiv 1 \pmod{p}$ that are not solutions to $x^2\equiv 1\pmod{p}$. The order of the group here is $p-1$, so the number of solutions to the first is a multiple of $\gcd(4,p-1)$; the number of solutions of the second is a multiple of $2$, and we know the two: $1$ and $-1$. Since every solution to $x^4\equiv 1\pmod{p}$ is a solution to $x^2\equiv 1 \pmod{p}$, there are solutions to $x^2\equiv -1\pmod{p}$ if and only if $\gcd(4,p-1) = 4$.