Doubt on showing that when $p$ is prime, the only unipotent class is $p-1$?

Question

I am trying to understand the solution of the following exercise:

A number $a$ is unipotent if $aneq1$ and $a^2equiv 1 pmod{p}$.

Show that when $p$ is prime, the only unipotent class is $p-1$.

The answer is:

$(p-1)xequiv1 pmod{p}$ has a unique solution in $Bbb{Z}/pBbb{Z}$. This solution is $xequiv -1 pmod{p}$. It is the same as $x=p-1$.

I may be doing some very silly mistake: We want to prove that $a=p-1$ is the only number such that $a^2equiv 1 pmod{p}$ but why does the above proof proves it? Aren't we proving only that $a$ is unipotent without - somehow - checking it for $(p-1),(p-2),(p-3),dots$?

shibai · Accepted Answer

Frankly I'm not quite sure what the answer's proof is trying to say, but the easiest way to see this result is to note that $Bbb Z/pBbb Z$ is a field, and thus the polynomial $x^2-1$ can only have at most two roots (i.e., as many roots as its degree). As $x=pm1$ both solve this polynomial, these are the only roots. As we're excluding $x=1$, this shows that there is a unique unipotent element modulo $p$.
That being said, in the case $p=2$ there are actually no unipotent elements since $1=-1mod 2$ is the only element that squares to $1$ in $Bbb Z/2Bbb Z$.

Doubt on showing that when $p$ is prime, the only unipotent class is $p-1$?

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