Let me add a few remarks concerning 2. If $p \equiv 3 \bmod 4$, then ${\mathbb F}_p(i) = {\mathbb F}_{p^2}$. The relative norm of $x+iy$ is the product of $x+iy$ and its conjugate $x-iy$, but the latter is the image of the Frobenius automorphism, i.e., $x-iy = (x+iy)^p$. This shows that $x^2 + y^2 = (x+iy)(x-iy) = (x+iy)^p$.

If $x+iy$ is an element with norm $-1$, i.e., with $x^2 + y^2 = -1$ in ${\mathbb F}_p$, then you get all other elements with norm $-1$ by multiplying $x+iy$ by an element with norm $1$. By Hilbert's Theorem 90, such elements have the form $\frac{c+di}{c-di}$; if $d = 0$, this quotient is $1$, so you may assume $d \ne 0$ and cancel $d$; writing $u = c/d$ then proves the second claim.

Let me add a few remarks concerning 2. If $p \equiv 3 \bmod 4$, then ${\mathbb F}_p(i) = {\mathbb F}_{p^2}$. The relative norm of $x+iy$ is the product of $x+iy$ and its conjugate $x-iy$, but the latter is the image of the Frobenius automorphism, i.e., $x-iy = (x+iy)^p$. This shows that $x^2 + y^2 = (x+iy)(x-iy) = (x+iy)^{p+1}$.

If $x+iy$ is an element with norm $-1$, i.e., with $x^2 + y^2 = -1$ in ${\mathbb F}_p$, then you get all other elements with norm $-1$ by multiplying $x+iy$ by an element with norm $1$. By Hilbert's Theorem 90, such elements have the form $\frac{c+di}{c-di}$; if $d = 0$, this quotient is $1$, so you may assume $d \ne 0$ and cancel $d$; writing $u = c/d$ then proves the second claim.