<s>[INST] Find the number of different residues mod $p$ of $(x^2+y^2)^2$ where $(x,p)=(y,p)=1$ Let $p=4k+3$ be a prime number. Find the number of different residues mod $p$ of $$(x^2+y^2)^2$$ where $(x,p)=(y,p)=1$
This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?
I think the answer is $\dfrac{p-1}{2}$.But How to prove it?
 [/INST] Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 \not\equiv 0 \pmod p$. As such, the number of possible residues of
$$(x^2+y^2)^2 \tag{1}\label{eq1}$$
is $\le \frac{p - 1}{2}$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 \le y \le \frac{p - 1}{2}$. In those cases, \eqref{eq1} becomes $4y^4$. To confirm they're all unique, let some $1 \le z \le \frac{p - 1}{2}$, where $z \neq y$, be such that
$$4y^4 \equiv 4z^4 \pmod p \; \Rightarrow \left(y^2 - z^2\right)\left(y^2 + z^2\right) \equiv 0 \pmod p \tag{2}\label{eq2}$$
Since $y^2 + z^2 \not\equiv 0 \pmod p$, this means that $y^2 - z^2 \equiv 0 \pmod p$. However, since the $\frac{p-1}{2}$ residues of values between $1$ and $\frac{p-1}{2}$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $\frac{p-1}{2}$.
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