Is the number of quadratic nonresidues modulo $p^2$, greater than the number of quadratic residues modulo $p^2$?

Question

Let $p$ be a prime. The number of quadratic nonresidues modulo $p^2$, is greater than the number of quadratic residues modulo $p^2$.

Is that statement true or false? Why?

Thank you.

Answer 1

Let's count the number of squares modulo $p^2$. I'll assume that $p$ is odd. The case $p=2$ is left as an exercise...

If $x^2$ is a multiple of $p$, then $x$ is a multiple of $p$, and $x^2 \equiv 0 \bmod {p^2}$.

If $x^2$ is a not multiple of $p$, then $x$ is a not multiple of $p$, and $x^2$ is a unit modulo $p^2$. There are $\phi(p^2)=p^2-p$ such units. Since there is a primitive root modulo $p^2$, half of the units are squares. Indeed, if $g$ is a primitive root, then all units are of the form $g^k$ and the squares are of the form $g^{2k}$.

So, the number of squares is $(p^2-p)/2+1$ and the number of nonsquares is $(p^2+p)/2-1$.

Alternatively, consider the map $x \mapsto x^2$ in the unit group. This map is a homomorphism and its image is the set of nonzero squares. The kernel is the set of solutions $x^2=1$, which is $\{\pm 1\}$. (There is something to prove here!) Hence the image has size $(p^2-p)/2$.