Proving that an integral domain has at most two elements that satisfy the equation $x^2 = 1$.

I like to be thorough, but if you feel confident you can skip the first paragraph.

Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds. If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.

We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).

Here's what I did:

For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$. Assume $a^2 = 1$. We want to show that letting $b^2 = 1$ as well will lead to a contradiction. So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution? If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction. If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction. Similarly, $ab = b$ won't work either. So it must be that $ab = c$. So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.

So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$. We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution. From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$. So $abc$ is not a new solution; it's just one.

At this point I'm stuck. I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$. What am I missing?

Hint: $x^2 - 1 = (x-1)(x+1)$. If this is $0 \ldots$

More generally, every element $\ell\in R$ of an integral domain $R$ cannot have more than two square roots. To see this, let $a$ be such that $a^2=\ell$, and suppose $b^2=\ell$ also for some $b\in R$. Then we can subtract one from the other and factor as $(a-b)(a+b)=0$, and deduce $b=\pm a$ via integrality.

You’ve shown that if $R$ has two distinct elements other than $1$ whose squares are $1$, then their product is a third such element. But in fact $R$ can’t have two distinct elements other than $1$ whose squares are $1$ in the first place. To see this, show that $x^2=1$ can have at most two solutions by factoring $x^2-1$ and using the fact that $R$ has no zero-divisors.