There is a sequence of three exercise in Ireland and Rosen's Introduction to Modern Number Theory, Chapter 8, page 106. I can do the first two, but can't finish the third. I can include the proofs to the first two if they are wanted.
Suppose that $p\equiv 1\pmod{4}$, $\chi$ is a character of order $4$, and $\rho$ is a character of order $2$. Let $N$ be the number of solutions to $x^4+y^4=1$ in $F_p$. Show that $N=p+1-\delta_4(-1)4+2\mathrm{Re} J(\chi,\chi)+4\mathrm{Re} J(\chi,\rho)$. (Here $J$ is the Jacobi sum, and $\delta_4(-1)$ is $1$ is $-1$ is a fourth power, and $0$ otherwise.) (Solved.)
By Exercise 7, $J(\chi,\chi)=\chi(-1)J(\chi,\rho)$. Let $\pi=-J(\chi,\rho)$. Show that $N=p-3-6\mathrm{Re}\pi$ if $p\equiv 1\pmod{8}$ and $N=p+1-2\mathrm{Re}\pi$ if $p\equiv 5\pmod{8}$. (Solved.)
Let $\pi=a+bi$. One can show (See Chapter 11, Section 5) that $a$ is odd, $b$ is even, and $a\equiv 1\pmod{4}$ if $4\mid b$ and $a\equiv -1\pmod{4}$ if $4\nmid b$. Let $p=A^2+B^2$ and fix $A$ by requiring that $A\equiv 1\pmod{4}$. Then show that $N=p-3-6A$ if $p\equiv 1\pmod{8}$ and $N=p+1+2A$ if $p\equiv 5\pmod{8}$.
My thoughts so far: I know I can also express $\pi=-J(\chi,\rho)=-\chi(-1)J(\chi,\chi)$.
I see that $\pi\in\mathbb{Z}[i]$, and that $\Re(\pi)^2+\Im(\pi)^2=|\pi|^2=|-\chi(-1)J(\chi,\chi)|^2=p$. So I can express $$ p=A^2+B^2=\Re(\pi)^2+\Im(\pi)^2. $$ If I fix $A$ by requiring that $A\equiv 1\pmod{4}$, is there a way to conclude that $A=\Re(\pi)$ when $p\equiv 1\pmod{8}$ and $A=-\Re(\pi)$ when $p\equiv 5\pmod{8}$ to get the desired result? I believe if $p\equiv 1\pmod{8}$ implies $4\mid b$, then $a\equiv 1\pmod{4}$, i.e., $A=\Re(\pi)$, and if $p\equiv 5\pmod{8}$ implies $b\nmid 4$, then $a\equiv -1\pmod{4}$, or $-a\equiv 1\pmod{4}$, i.e., $-\Re(\pi)=A$, which is precisely what I want, but don't see how to get there.
If it's any help, I know that $-1$ is a fourth power iff $p\equiv 1\pmod{8}$, which tells me $\pi=-J(\chi,\chi)$ when $p\equiv 1\pmod{8}$, and $\pi=J(\chi,\chi)$ when $p\equiv 5\pmod{8}$. Then $|J(\chi,\chi)|=\Re(J)^2+\Im(J)^2=p$, but $\Re(\pi)=-\Re(J)$ when $p\equiv 1\pmod{8}$ and $\Re(\pi)=\Re(J)$ when $p\equiv 5\pmod{8}$, which gives me the opposite sign of what I want.
Thanks for any help.