In a numerical experiment I notice for sum moduli $N$ there are much less than $N/2$ perfect squares. I had chosen a large number, the simplest example is $N=8$.

Using the Chinese Remainder Theorem (as here) and prime factorization one can solve

$$x^2 \equiv a \mod pq $$

in terms of two equations $x^2 \equiv a \mod p $ and $x^2 \equiv a \mod q $.

If $a$ is a quadratic residue with respect to all of these moduli you can get a solution or you get none at all.

The number of quadratic residues mod $n$ is OEIS sequence A000224:

a(n) = 1, 2, 2, 2, 3, 4, 4, 3, 4, 6, 6, 4, 7, 8, 6, 4, 9, 8, 10, 6

This function is multiplicative and there is a closed formula for each prime power modulus:

$$ a(p^e) = \begin{cases} \big[\frac{1}{6} p^e\big] & \text{ if } p=2 \\ \big[\frac{1}{2(p+1)} p^{e+1}\big] & \text{ if } p\neq2 \end{cases} $$

How to prove this result? Why does the floor function play a role here?


The "floor" is actually hiding what is actually happening. You should see Stangl's paper for the proof (MAA link) (PDF link).

The original formula is (for $n \ge 3$):

$$ s(p^n) = \begin{cases} \frac{p^{n+1}+p+2}{2p + 2} & p\neq 2 \text{ and $n$ even} \\ \frac{p^{n+1}+2p+1}{2p + 2} & p\neq 2 \text{ and $n$ odd} \\ \frac{2^{n-1}+4}{3} & p = 2 \text{ and $n$ even} \\ \frac{2^{n-1}+5}{3} & p = 2 \text{ and $n$ odd} \end{cases}$$

which is proven using the recurrence formula:

$$ s(p^n) = q(p^n) + s(p^{n-2}) $$

where $s(x)$ is the number of squares and $q(x)$ is the number of quadratic residues (only units).


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