Let $p$ be a prime such that $p \equiv 1 \mod 8$. Then we know there exists $a,b \in \mathbb{Z}$ such that $p = a^2 + 2b^2$. But at the same time $p \equiv 1 \mod 4$, so there also exists $c,d \in \mathbb{Z}$ such that $p = c^2 + d^2$.
My question is: Are there any interesting connections between $a$ and $c$ (or $b$ and $d$)?
Here is another example I have in mind: Let $p \equiv 1 \mod 6$ and $\chi_6$ be a primitive character $\mod p$ of order $6$. Let $g$ be a primitive root modulo $p$ and $Z$ denote the index of $2$ with respect to $g$ modulo $p$ (that is, $g^Z \equiv 2 \mod p$). Consider the Jacobi sum \begin{align*} J(\chi) = \sum_{a = 1}^{p-1} \chi_6(a) \chi_6(a-1). \end{align*} Then we know that $J(\chi_6) = \left(\frac{-1}{p}\right)\frac{1}{2}(u + v\sqrt{-3})$, where $\left(\frac{\cdot}{p}\right)$ is the Legendre symbol, $u,v \in \mathbb{Z}$ such that $4p = u^2 + 3v^2$, $u \equiv 1 \mod 3$, $v \equiv Z \mod 3$, and $3v \equiv (2g^{(p-1)/3}+1)u \mod p$.
On the other hand, $\chi_3 = \chi_6^2$ is a character of order $3$, and $J(\chi_3) = \left(\frac{-1}{p}\right)\frac{1}{2}(r + s\sqrt{-3})$, where $r,s \in \mathbb{Z}$ such that $4p = r^2 + 3s^2$, $r \equiv 1 \mod 3$, $s \equiv 0 \mod 3$, and $3s \equiv (2g^{(p-1)/3}+1)r \mod p$.
We can show that \begin{equation} (\star) \,\,\,\, \left(J(\chi_3^2)\right)^3 = \left((-1)^{(p-1)/6}J(\chi_6^5)\right)^3 \end{equation}
and because $J(\chi_6^5) \equiv (-1)^{(p-1)/6 + 1}u \mod p$ and $J(\chi_3^2) \equiv r \mod p$, we get $u^3 \equiv r^3 \mod p$.
Going back to the case where $p \equiv 1 \mod 8$, with $\chi_8$ a primitive character $\mod p$ of order $8$ and $\chi_4 = \chi_8^2$, I am wondering if there is a similar congruence to $(\star)$ for $J(\chi_8)$ and $J(\chi_4)$, whose values involve $a,b,c,d$ defined above analogous to $u,v,r,s$.
