Counting pairs of squares in ${\mathbb Z}_n$ with certain distance Let $S_n = \{ x^2 \pmod{n} \mid x \in \mathbb Z \}$ denote the set of squares in ${\mathbb Z}_n$.
Define $S_n(d) = \{ (x, y) \in S_n^2 \mid x + d \equiv y \pmod{n} \}$.
Is there an explicit formula for $|S_n(d)|$ ?

Update:
As mentioned in comments, if $d \equiv 0 \pmod{n}$, then $|S_n(d)| = |S_n|$ and there is a formula for it according to Walter D. Stangl's paper "Counting Squares in ${\mathbb Z}_n$" (MAA link) (PDF link).
I am still looking for a general result where $d \not\equiv 0 \pmod{n}$.
 A: This is far from being a complete answer but it might give you some clue to follow up. What I give is a closed form when $p$ is an odd prime and $p\nmid d$:
\begin{align} S_p(d) &= \frac{ p+1+(\tfrac{-d}{p}) + (\tfrac{d}p)}4 \\&= \frac{ p+1 +(1+(-1)^{\frac{p-1}2})(\tfrac{d}p)}4\end{align}
$(\frac dp)$ is the Legendre symbol which is zero if $p$ divides  $d$, $+1$ when $d$ is a square in $\mathbb Z_p$ and $-1$ otherwise. I'm using the elementary theory of quadratic residues and non residues and Legendre symbols freely in the rest of the answer. (by the way zero is neither a quadratic residue nor a quadratic non-residue which is usually a source of some confusion)
The method of proof goes back to Gauss, first I'm using the following result, (I give an sketch of the proof at the end):
$$ \sum_{t=1}^p \left(\frac{t(t+d)}p\right) = -1\tag{1}\label{eq1}$$
Call $RR$ the number of integers $t$ mod $p$ such that both $t$ and $t+d$ are both quadratic residues, in the same way let $RN$ the number of integers $t$ mod $p$ such that both $t$ is a quaratic residue and $t+d$ a quadratic non-reside, and define $NR$ and $NN$ in a similar way. Then the sum in \eqref{eq1} can be written 
    $$ RR + RN + NR + NN = -1 \tag{2}\label{eq2} $$
now  observe that the sum $RR+RN$ counts once every quadratic residue except $-d$ if it is a quadratic residue, as there are $\tfrac{p-1}2$ quadratic residues and ${1+(\tfrac{-d}p)}2 =1$ exactly when $-d$ is a quadratic residue then
$$ RR + RN = \frac{p-1}2 - \frac {1+(\tfrac{-d}p)}2 = \frac{p-2-(\tfrac{-d}p)}2\tag{3}\label{eq3}$$ 
With similar arguments we find:
$$ NR+NN=\frac{p-2+(\tfrac{-d}p)}2\tag{4}\label{eq4}$$
$$ RR+NR=\frac{p-2-(\tfrac{d}p)}2\tag{5}\label{eq5}$$
$$ RN+NN=\frac{p-2+(\tfrac{d}p)}2\tag{6}\label{eq6}$$
Summing \eqref{eq2}, \eqref{eq3} and \eqref{eq4} we get after simplifying
$$ 2RR +2NN = p-3$$
summing \eqref{eq3} and \eqref{eq4} and substracting \eqref{eq5} and \eqref{eq6} we get
$$ 2RR - 2NN = -(\tfrac{-d}p)-(\tfrac dp) $$
so we get
$$ 4RR = p-3 -(\tfrac{-d}p)-(\tfrac{d}p) $$
Now we have 
    $$ S_p(d) = RR + \tfrac 12 \bigl(1 -(\tfrac{-d}p)\bigr)+ \tfrac 12 \bigl(1 -(\tfrac{d}p)\bigr) $$
and combining both we get the claim.
Now we prove \eqref{eq1}. Call the left hand sum $T(d)$ then 
\begin{align}
 T(d) &= \sum_{t=1}^p \left(\frac{t(t+d)}p\right) \\
      &= \sum_{t=1}^p \left(\frac{dt(dt+d)}p\right)=\sum_{dt=1}^p \left(\frac{d^2}p\right)\left(\frac{t(t+1)}p\right)\\
         &= \sum_{t=1}^p \left(\frac{t(t+1)}p\right) \\
         &= T(1) 
\end{align}
Now we can tke the sum
$$ T(0)+T(1)+\dots+T(p-1) = \sum_{t=1}^p \left(\frac tp\right) \sum_{d=0}^p\left(\frac{(t+d)}p\right) = 0 $$
but $T(0) = p-1$ and $T(1)=T(2)=\dots =T(p-1)$ so
    $$  T(0)+T(1)+\dots+T(p-1)  = p-1 + (p-1)T(1)  = 0 $$
and we get \eqref{eq1}.
If you want to extend this to prime powers you could try to reproduce this argument for prime powers using a primitive root $g$ mod $p^k$, if $t \equiv g^a$ define $\chi(t)= (-1)^a$, and try to evaluate $\sum \chi(t(t+d))$.
