$p=3 \pmod 4$ is a prime. Let $b\in \mathbb F_p^*$.
Show that the equation $v^2=u^4-4b$ has $p-1$ solutions $(u,v)$ with $u, v \in \mathbb F_p$.
If we write the given equation as $v+u^2=x$ and $v-u^2=-4by$ where $y$ is the inverse of $x$ in $\mathbb F_p^*$ and solve, we get $u^2=-\frac {4b+x^2} {2x}$. So, I need to check when this quantity $-\frac {4b+x^2} {2x}$ will be a perfect square in $\mathbb F_p$. Can anyone please help me with this?
Let us introduce a "new" variable $w=u^2$. Shiva's idea shows that there are $p-1$ pairs $(v,w)$ such that $v^2=w^2-4b$ - one for each non-zero value of $x$.
The claim follows from the fact that $\gcd(p-1,4)=2$. A consequence of this is that the set of fourth powers in $\mathbb{F}_p$ is exactly the same as the set of squares (the group is cyclic!). Furthermore, both functions $f(x)=x^2$ and $g(x)=x^4$ are 2-1. This follows either by a counting argument or from the observation that $f$ and $g$ are both endomorphisms of $\mathbb{F}_p^*$ with kernel $\{\pm1\}$. So the number of solutions of $v^2=w^2-4b$ equals the number of solutions of $v^2=u^4-4b$, and the former we know to be equal to $p-1$.
An alternative way of seeing the same thing (every square is a fourth power) is by Little Fermat. If $w$ is non-zero, then $$ w^2=w^2\cdot1=w^2\cdot w^{p-1}=w^{p+1}=(w^{(p+1)/4})^4. $$ Here $p\equiv 3\pmod4$ implies that $(p+1)/4$ is an integer.