# In an Integral Domain, if $a^2=1$, then $a=1$ or $a=-1$

I can’t think of a way to prove this, can anyone help?

EDIT: I noticed this is a really simple question, and the confusion I made came from not seeing what was false intuition and what was the real Algebraic structure I was studying, I’m just starting to learn Algebra. Thanks for the answers, and sorry for those who thought my question didn’t follow the standards it should have, I really didn’t mean to cause this negative impact.

#### Solutions Collecting From Web of "In an Integral Domain, if $a^2=1$, then $a=1$ or $a=-1$"

$a^2-1=0$ is equivalent to $a^2-1=(a-1)(a+1)=0$ since the domain is integral, $a=1$ or $a=-1$.

For added insight, let’s do a slightly more general case. We prove that a nonzero quadratic polynomial $f(x)$ over a domain $D$ has at most two roots $\,\color{#c00}{a\neq b},\,$ by using the Factor Theorem twice, and using that $D$ is a domain, so a product of nonzero elements remains nonzero.

$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in D[x]\\ \rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$

Comparing degree shows $h(x) = c$ is constant, and $f\neq 0\,\Rightarrow\,c\neq 0.\,$ If $f$ had a third root $r$ then $(r\!-\!b)(r\!-\!a)c = 0$, contra each factor is nonzero hence so is their product, since $D$ is a domain.

Remark  The proof immediately generalizes by induction to yield that a nonzero polynomial of degree $n$ has at most $n$ roots over a domain. In fact this is a characteristic property of domains, since if $R$ is not a domain then there are $a,b\neq 0$ with $\,ab = 0\,$ so $\,f = ax\,$ has $2$ roots, $\,x = 0,b.$