<s>[INST] If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?

I notice quadratic residues of odd prime $13$ : 
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv 4\\3^2\equiv 10^2\equiv 9\\4^2\equiv 9^2\equiv 3\\5^2\equiv 8^2\equiv 12\\6^2\equiv 7^2\equiv 10$$
$4^{-1}=10\\9^{-1}=3\\1^{-1}=1\\12^{-1}=12$ 
Is this true always ? How to go about proving ? thanks! 
 [/INST] If $a^2 = b$ in a ring $R$ with unit, $a$ is invertible if and only if $b$ is and then $b^{-1} = (a^{-1})^2$. (Tell if this is clear to you, if not, I'll detail in an edit.) Apply this to $R = \mathbf{Z} / n \mathbf{Z}$. Apply the latter to the case $n = p$ is prime. ;-)
Detail. If $a$ is (for instance left) invertible of left inverse $c$ then $ca = 1$ so that $ca^2 = a$ which implies that $c^2 a^2 = ca = 1$, that is $c^2 b = 1$ and $b$ is invertible. If $b$ is (let's say left) invertible then $db=1$ and then $(d a) a = 1$ and $a$ is left invertible. Same for right inverse. The relation $c^2 a^2 = 1$ shows you the assertion on quadratic residue, that is, that if you are invertible and a quadratic residue, that you your inverse is also a quadratic residue.
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