= Solution
Suppose $x^2\equiv-1\pmod p$. Then $x\ne0$ modulo the <prime number> $p$, so <Fermat's little theorem> gives $x^{p-1}\equiv1\pmod p$. But $p=4m+3$ makes $(p-1)/2=2m+1$ odd, and therefore
$$
x^{p-1}=(x^2)^{(p-1)/2}\equiv(-1)^{2m+1}=-1\pmod p.
$$
This contradicts $1\not\equiv-1\pmod p$, since $p$ is odd. Thus \b[$-1$ 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 $\boxed{p^2}$.
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