How to solve $x^2 \equiv [1]$ in $\Bbb Z_5$

Question

I would like to know how to solve $x^2 \equiv [1]\text{ in }\Bbb Z_5$?

How to solve this kind of equation in general?

Answer 1

There are five values for $x$: $[0],[1],[2],[3],[4]$.
Square them, and you get $[0],[1],[4],[9],[16]$, which are $[0],[1],[4],[4],[1]$.
You get $[1]$ twice, so the answers are $x=[1]$ and $x=[4]$.

For general $n$, $x^2=[1]$ means that $x^2-1$ is a multiple of $n$.
Ordinary algebra says that $x^2-1=(x-1)(x+1)$.
If $n$ is prime, then either $x-1$ is a multiple of $n$, so $x=[1]$ in $\mathbb{Z}_n$, or $x+1$ is a multiple of $n$, so $x=[n-1]$ in $\mathbb{Z}_n$.

There can be more solutions if $n$ is not prime, For example $n=8$, there are four solutions $x=[1],[3],[5],[7]$. Things are better behaved if $n$ is prime.

Answer 2

Modulo $5$, the only possible remainders are $0,\pm1$ and $\pm2$. Squaring them, we have $0,1$ and $4$. In general, when working modulo $2n+1$, the remainders are $0,\pm1,\pm2,\ldots,\pm n$. This is especially useful when dealing with even powers. Hope this helps.