Let $p=4k+3$ be a prime number. Find the number of different residues mod p of $(x^{2}+y^{2})^{2}$ where $(x,p)=(y,p)=1.$
1. Let \( p = 4k + 3 \) be a prime number. We need to find the number of different residues modulo \( p \) of \( (x^2 + y^2)^2 \) where \( (x, p) = (y, p) = 1 \). This means \( x \) and \( y \) are nonzero modulo \( p \).

2. Consider the expression \( (x^2 + y^2)^2 \). We need to determine the possible values of this expression modulo \( p \).

3. First, note that since \( x \) and \( y \) are nonzero modulo \( p \), \( x^2 \) and \( y^2 \) are quadratic residues modulo \( p \). 

4. We can simplify the problem by considering the case where \( x = y \). In this case, we have:
   \[
   (x^2 + y^2)^2 = (x^2 + x^2)^2 = (2x^2)^2 = 4x^4
   \]

5. Since \( x \) is nonzero modulo \( p \), \( x^4 \) will cover all the quadratic residues modulo \( p \). This is because \( x^2 \) covers all quadratic residues, and squaring a quadratic residue will still be a quadratic residue.

6. Therefore, \( 4x^4 \) will cover all the quadratic residues multiplied by 4. Since 4 is a constant, it does not change the nature of the residues modulo \( p \).

7. The number of quadratic residues modulo \( p \) is \( \frac{p-1}{2} \). This is because there are \( p-1 \) nonzero elements modulo \( p \), and half of them are quadratic residues.

8. Hence, the number of different residues modulo \( p \) of \( (x^2 + y^2)^2 \) is the same as the number of quadratic residues modulo \( p \), which is \( \frac{p-1}{2} \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{p-1}{2} } \)