Let $p>2$ be a prime number. How many integers $a \in\{0,1, \cdots, p-1\}$ are there such that $c^{2}=$ $p k+a$ has a solution?
Let $a$ and $b$ be two different remainders modulo $p$. Suppose that $a^{2} \equiv b^{2}[p]$. Then $(a-b)(a+b) \equiv$ $0[p]$, but $p$ is prime. By Gauss's lemma, $p$ divides $a-b$ or $a+b$. Thus $a \equiv b$ or $-b[p]$.

Furthermore, $a^{2}$ and $(p-a)^{2}$ are congruent modulo $p$. Thus if $p \neq 2,0$ is a square and the remainders $1,2, \cdots, \frac{p-1}{2}$ have different squares. All squares are obtained this way. Therefore, there are $\frac{p+1}{2}$ squares modulo $p$ if $p \neq 2$, and modulo 2 it can be verified that there are 2 squares.