Show that $2a \equiv -(-1)^m {2m \choose m} \pmod{p}$ where $m={p-1 \over 4}$

Question

This is Problem $26$ from chapter $8$ of "A Classical Introduction to Modern Number Theory" by Ireland-Rosen :

Let $p$ be a prime, $p \equiv 1 \pmod{4}$, $\chi$ a multiplicative character of order $4$ on $F_p$, and $\rho$ the Legendre symbol. Put $J(\chi,\rho)=a+ib$. Show

(a) $N(y^2+x^4=1)=p-1+2a$

(b) $N(y^2=1-x^4)=p+\sum{\rho(1-x^4)}$

(c) $2a \equiv -(-1)^m {2m \choose m} \pmod{p}$ where $m={p-1 \over 4}$

(d) Verify (c) for $p=13,17,19$.

I can solve (a),(b),(d) but I'm stuck with (c). Using (a),(b) we have $2a-1 \equiv \sum{\rho(x^4-1)} \equiv \sum{\rho(x^2+1)\rho(x^2-1)} \pmod{p}$ since $p \equiv 1 \pmod{4}$. I have no idea how to move forward. Please help me out.

Answer 1

I'm adding this answer only as a future reference.

Following @metamorphy's comment, $$ \begin{align*} 2a-1 & \equiv \sum_x{\rho(1-x^4)} \equiv \sum_x(1-x^4)^{2m} \equiv \sum_x \sum_{i=1}^{2m}{{2m \choose i}(-x^4)^i} \\ & \equiv \sum_{i=1}^{2m} \sum_x{{2m \choose i}(-x^4)^i} \equiv \sum_x {2m \choose m}(-1)^m x^{p-1} + \sum_x (-1)^{2m} x^{2(p-1)} \\ & \equiv {2m \choose m}(-1)^m(p-1) + (p-1) \\ & \equiv -(-1)^m{2m \choose m}-1 \pmod{p} \end{align*} $$ So, $2a \equiv -(-1)^m{2m \choose m} \pmod{p}$.