On the proof of Weils hyperelliptic theorem

Question

Let $p$ be a large prime. Consider $F_p$ Theorem:

Let $P$ be an element in $F_p[x]$ of degree $k$, assume that $P$ is not a constant multiple of a square. Then the number of solutions $(x,y)$ in $(F_p)^2$ to $y^2 = P(x)$ is $p+O_k(\sqrt p)$

A proof of this result can be found in: https://terrytao.wordpress.com/2009/08/18/the-least-quadratic-nonresidue-and-the-square-root-barrier/#more-2664

The proof uses the polynomial method, defining the polynomial $Q(x) = P^l(R(x,x^p)+P^{(p-1)/2}S(x,x^p))$ for $l\cong \sqrt p$ and polys $R(x,z),S(x,z)$ satisfying some degree condition. Then one shows you can choose such $S,R$ so that for each $x$ with $P(x)$ a square, $Q$ vanishes at least $l$ times.

I don't understand why we need to multiply by $P^l$, why doesn't Terry's argument go through even without it?

Thanks

Answer 1

Yeah it's actually stupid:

We want the derivative to also be of the form with $P^((p-1)/2)$, and a way to promise that is multiply by $P^l$ and give copies of it when taking the derivative.