There is a field $\Bbb F$ such that the equation $x^2=1$ have more than two solutions (for some $x\in\Bbb F$)? 
There is a field $\Bbb F$ such that the equation $x^2=1$ have more than two solutions (for some $x\in\Bbb F$)?

This question comes suddenly to my mind. 
I know that if $\Bbb F=\Bbb C$ then the fundamental theorem of algebra states that the equation $x^2=1$ have two solutions, but this would imply that for an arbitrary field $x^2=1$ have, at most, two solutions?
Im sorry if this question is trivial but at this moment Im unable to achieve a conclusion from the axioms that define a field.
 A: No: over an integral domain, a polynomial of degree $n$ has at most $n$ roots.
This is a consequence of the fact that a polynomial $p$ is divisible by $X-\alpha$ if and only if $p(\alpha)=0$.
A: If $x^2=1$ then $(x-1)(x+1)=0$, and since $F$ is a field either $x-1=0$ or $x+1=0$.
In general, if $f(x)$ is a polynomial over a field $F$ of degree $n$, then $f$ has at most $n$ roots in $F$.
A: No, in any field $x^2 = 1$ iff $x^2-1 = 0$ iff $(x-1)(x+1) = 0$ and in a field (no zero-divisors) this means that either $x-1 = 0$ (so $x=1$) or $x+1 = 0$ (so $x=-1$).
A: Hint : $x^2 = 1 \iff (x-1)(x+1) = 0$ and a field is an integral domain 
A: If $\alpha$ solves
$x^2 = 1, \tag{1}$
so that
$\alpha^2 = 1, \tag{2}$
or
$\alpha^2 -1 = 0, \tag{3}$
we have
$(\alpha - 1)(\alpha + 1) = 0; \tag{4}$
if now
$\alpha \ne 1, \tag{5}$
so that
$\alpha -1 \ne 0, \tag{6}$
then since we are operating in a field,
$\alpha + 1 = 0, \tag{7}$
or
$\alpha = -1. \tag{8}$
At most two possibilities.  And that's all.
A: In any non zero ring $R$ with $1$, which has no nonzero zero divisors, $x^2=1$ has exactly $1$ root if $char(R)=2$ and exactly 2 roots if $char(R)\neq2$.
