It is clear that there is $ 2² \equiv 4 \mod 11 $ and $ 3² \equiv 9 \mod 11 $
But we have also for example: $7² = 49 \equiv 5 \mod 11$ ?
Every number from $ \mathbb{Z_{11}}$ is equivalent to some other number in $\mathbb{Z_{11}}$
Does it admit one solution for each number $a \in \mathbb{Z_{11}} $?
Thank you for your help.
$\mathbb Z_{11}$ is a field, and therefore each element can have at most two square roots.
Furthermore, if $x^2=a$, then $(-x)^2=a$ too, and $x\ne -x$ unless $x=0$.
This means that exactly half of the ten nonzero elements in $\mathbb Z_{11}$ are squares; you can find them as $$ 1^2, 2^2, 3^2, 4^2, 5^2 \pmod{11} $$