 Solution

Suppose $x^2\equiv -1(\mathrm{mod}p)$. Then $x\ne 0$ modulo the prime number $p$, so Fermat's little theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. But $p=4m+3$ makes $(p-1)/2=2m+1$ odd, and thereforeThis contradicts $1\not\equiv -1(\mathrm{mod}p)$, since $p$ is odd. Thus $-1$ is not a quadratic residue modulo $p$. Apply part (a) to the finite field $F={\mathbb{F}}_p$: the pairs $F^2$ with the given operations form a field, and there are $p$ choices for each coordinate, so its cardinality is $\boxed{p^2}$.