$\mathbf{\Sigma}-5$. Let $p$ be an odd prime, $Z_{p}$ denote the residue class field modulo $p$. How many elements does the following set contain
$$
\left\{x^{2}: x \in Z_{p}\right\} \cap\left\{y^{2}+1: y \in Z_{p}\right\} ?
$$
This intersection has $[(p+3) / 4]$ elements. First, consider the solution set of the following equation:
$$
x^{2}=y^{2}+1.
$$

Rewriting it as $(x+y)(x-y)=1$, we see that for any non-zero element $r$ in $Z_{p}$, there is exactly one solution to the above equation, i.e., $x+y=r, x-y=r^{-1}$ or $x=\left(\frac{p+1}{2}\right)(r + r^{-1}), y=\left(\frac{p+1}{2}\right)\left(r-r^{-1}\right)$. Therefore, equation (*) has $p-1$ solutions. (Note: Here we are using operations in $Z_{p}$. For example, when $p=5$, $Z_{p}=\{0,1,2,3,4\}, 3+4=2, 3^{2}=4$, $2 \times 3=1, 2^{-1}=3$, etc. — detailed notes)

On the other hand, the elements in the intersection $x^{2}=y^{2}+1$ can also be formed by the pairs $(x,-y), (-x, y), (-x, -y)$, and $(x, y)$, which are 4 different pairs, except for the case where $x=0$ and $y=0$. The first two pairs are distinct. The element 1 is generated by (1,0) and (-1,0). When 0 belongs to the intersection, we set $c=1$; otherwise, we set $c=0$. Thus, the intersection contains $1+c+d$ elements. From the previous paragraph, we know that $p-1=2+2c+4d$.

It is easy to see that $c=1$ if and only if $p-1$ is divisible by 4. Solving for $d$ in each case so that $1+c+d=[(p+3) / 4]$.