Using Wilson's Theorem to show a particular congruence holds.

Question

I am trying to answer the following: Does the congruence $x^2 \equiv -1$ (mod $p$) have any solutions if $p \equiv 3$ (mod $4$)? If so, how many incongruent solutions does it have? If not, why not?

I know from the previous part of the question that if $p$ is a prime and $p \equiv 1$ (mod $4$), then the congruence $x^2 \equiv -1$ (mod $p$) has two incongruent solutions, namely $x \equiv \pm (\dfrac{p-1}{2})!$ (mod $p$).

I am completely unsure how to even approach solving this problem. Any hints would be appreciated.

Answer 1

André Nicolas has an approach that uses Wilson's theorem, as requested by the OP, but I guess I'll leave my answer up.

---

No, it has no solutions. The first step is to observe that, for an odd prime $p$, $$x^2\equiv -1\bmod p\iff x\text{ has order 4 in }(\mathbb{Z}/p\mathbb{Z})^\times$$ But the group $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic, so $$\text{there exists an element of order $d$ in }(\mathbb{Z}/p\mathbb{Z})^\times\iff d\text{ divides }\lvert(\mathbb{Z}/p\mathbb{Z})^\times\rvert=p-1$$ Thus, for an odd prime $p$, there exist solutions to $x^2\equiv -1\bmod p$ if and only if $4$ divides $p-1$, i.e., if and only if $p\equiv 1\bmod 4$.

Thus, there are no solutions when $p\equiv 3\bmod 4$.

Answer 2

The following is a proof that uses Wilson's Theorem. There are "easier" (and group-theoretically more natural) proofs that do not use Wilson's Theorem. The idea is due to Dirichlet.

Let $p$ be a prime of the form $4k+3$. We will assume that the congruence $x^2\equiv -1\pmod{p}$ has a solution, and use Wilson's Theorem to derive a contradiction.

If $x^2\equiv -1\pmod{p}$ had a solution $c$, then it would have exactly $2$ solutions, namely $c$ and $p-c$.

If $1\le a\le p-1$ and $1\le b\le p-1$, with $a\ne b$, call $a$ and $b$ buddies if $ab\equiv -1\pmod{p}$. Apart from $c$ and $p-c$, all numbers $a$ in the interval $1\le a\le p-1$ has a unique buddy $b\ne a$ in the interval $1\le b\le p-1$. So the numbers in the interval are divided into $\frac{p-3}{2}$ pairs of buddies, plus the numbers $c$ and $p-c$.

It follows that $$(p-1)!\equiv (-1)^{(p-3)/2}(c)(p-c)\equiv 1\pmod{p}.$$ This contradicts Wilson's Theorem. It follows that there cannot be a $c$ such that $c^2\equiv -1\pmod{p}$.

Remark: The OP alluded to a Wilson's Theorem proof of the fact that if $p$ is of the form $4k+1$, then the congruence $x^2\equiv -1\pmod{p}$ has a solution. We give a different Wilson's Theorem based proof, that uses the ideas in the answer above.

Suppose to the contrary that $p=4k+1$ and the congruence $x^2\equiv -1\pmod{p}$ has no solution. Then all the numbers from $1$ to $p-1$ are divided into buddy pairs. It follows that $(p-1)!\equiv (-1)^{2k}\equiv 1\pmod{p}$, contradicting Wilson's Theorem.

Answer 3

We assume $p$ is an odd prime. You know that $$\,p\equiv 1\pmod{\! 4}\,\Rightarrow\, (x^2\equiv -1\pmod{\! p}\,\text{ is solvable})$$ by your constructive proof, namely $x\equiv\pm\left(\frac{p-1}{2}\right)!\pmod{\! p}$ works as a solution. Proofs of this have been discussed here.

Now you want to prove it in the other direction, namely $$(x^2\equiv -1\!\pmod{\! p} \,\text{ is solvable})\,\Rightarrow\,p\equiv 1\pmod{\! 4}$$

For two proofs, see this answer with $(a,b)=(x,1)$, where it's proved that more generally:

$p\nmid a,b$ and $p\mid a^2+b^2$ implies $p\equiv 1\pmod{\! 4}$.

You can show $\iff$ at once by proving Euler's criterion (I try to prove it simply in this answer, or you can see this question for combinatorial proofs, in which André Nicolas extends his combinatorial argument in your question to prove a more general statement).