Let $p\equiv 3\pmod 4$ and $G$ be the set of nonzero quadratic residues modulo $p$ (so $G=(p-1)/2$). For $1\leq a\leq p-1$, let $G\_a=\{(a+g)\pmod p\mid g\in G\}$. What is the size of $G\_0\cap G\_a$?


I tested for $p=3,7$ and $11$, and the size of the intersection is always $(p-3)/4$. Is this always true? If so, it sounds like it should be a basic fact of quadratic residues, but I couldn't find it.



Shifting quadratic residues
Yes, the intersection has $(p-3)/4$ elements and this is standard, probably this specific claim also is written somewhere. We have $$\left|G\_0\cap G\_a\right|=\sum\_{x} \chi\_{G\_0}(x)\cdot \chi\_{G\_0}(x-a)=\sum\_x \frac{1-(\frac{x}p)-\delta(x)}2\cdot \frac{1-(\frac{x-a}p)-\delta(x-a)}2.$$
Expand the brackets and calculate. $\sum 1=p$, $\sum (\frac{x}p)=\sum (\frac{x-a}p)=0$, $-\sum \delta(x)=-\sum \delta(x-a)=-1$, $\sum \delta(x)\delta(x-a)=0$, $\sum \delta(x)(\frac{x-a}p)=(\frac{-a}p)$, $\sum \delta(x-a)(\frac{x}p)=(\frac{a}p)=-(\frac{-a}p)$. Finally $\sum (\frac{x}p)(\frac{x-a}p)=\sum\_{x\ne 0}(\frac{x}p)(\frac{x-a}p)=\sum\_{x\ne 0} (\frac{x^{-1}}p)(\frac{x-a}p)=\sum\_{x\ne 0}(\frac{1-ax^{-1}}p)=\sum\_{t\ne 1}(\frac{t}p)=-1$. Totally we get $(p-3)/4$.


