Prove that if $p$ is prime, then an element of $\mathbb{Z} /p \mathbb{Z}$ has at most two square roots.

Question

We say that $[a]$ is a square root of $[b]$ in $\mathbb{Z}/p\mathbb{Z}$ if $[a]^2= [b]$. Prove that if $p$ is prime, then an element of $\mathbb{Z}/p \mathbb{Z}$ has at most two square roots?

Answer 1

If $\Bbb F$ is any field, and for

$a, b \in \Bbb F \tag 1$

we have

$a^2 = b^2, \tag 2$

then

$(a + b)(a - b) = a^2 - b^2 = 0; \tag 2$

if then

$a \ne b, \tag 3$

we have

$a - b \ne 0, \tag 4$

whence

$a + b = 0, \tag 5$

that is

$b = -a; \tag 6$

these considerations reveal that if $a^2$ has more than one square root, i.e. if (4) binds, than the only candidates for $\sqrt{a^2}$ are $a$ and $-a$; i.e., $\sqrt{a^2}$ has at most two possible values.

Of course with $p$ a prime, $\Bbb Z/p\Bbb Z$ is a field and the above argument takes hold.