This is Exercise 10.17a from Silverman's $\textit{The Arithmetic of Elliptic Curves}$.
Let $p \equiv 3 \, (\text{mod 4})$ be a prime and let $D \in \mathbb{F}_p^{\times}.$
Show directly that the equation $$C: v^2 = u^4 - 4D$$ has $p-1$ solutions $(u,v) \in \mathbb{F}_p \times \mathbb{F}_p.$
Hint: Since $p \equiv 3 \, (\text{mod 4})$, the map $u^2 \mapsto u^4$ is an automorphism of $(\mathbb{F}_p^{\times})^2$. (This is true due to the fact that $\left(\frac{-1}{p}\right) = -1$ for $p \equiv 3 \, (\text{mod 4})$ a prime.)
I've been stuck on this problem for longer than I feel like I should. The directions indicate to "show directly," but I feel like there are too many different cases to consider to do this efficiently. The only "insight" I've gotten so far is that if $-4D$ is a perfect square, then $(0,\pm \sqrt{-4D})$ are two solutions. If $-4D$ isn't a perfect square, then by the Hint, $4D$ is a perfect square and so $(\pm \sqrt[4]{4D}, 0)$ are two solutions. But this leaves $p-3$ solutions undetected (hooray for the case $p = 3$).
My main issues are how to "show directly" there are $p-1$ solutions and where the Hint is mainly applied. I seem to have ideas that are indirect with no plans on how to tackle them directly.
The point is that, following the hint, the number of solutions of $$ v^2=u^4-4D\qquad(1) $$ is equal to the number of solutions of $$ v^2=u^2-4D.\qquad(2) $$ But equation $(2)$ can be rewritten to read $$ 4D=u^2-v^2=(u-v)(u+v). $$ Because $4D\neq0$ there are $p-1$ pairs $(x,y)\in\Bbb{F}_p^*\times\Bbb{F}_p^*$ such that $xy=4D$. Because the determinant of the linear system $$ \left\{\begin{array}{ccc}u+v&=&x\\u-v&=&y\end{array}\right. $$ equals $-2\neq0$, it has a unique solution $(u,v)$ to each given $(x,y)$. The claim follows.