# On the sum $\sum_{x=0}^{(p-1)/2}(\frac{x^{4n}+cx^{2n}+d}p)$ with $p$ an odd prime

Let $$p$$ be an odd prime, and let $$n$$ be a positive integer. For $$c,d\in\mathbb Z$$ we define $$F_p^{(n)}(c,d):=\sum_{x=0}^{(p-1)/2}\left(\frac{x^{4n}+cx^{2n}+d}p\right),$$ where $$(\frac{\cdot}p)$$ is the Legendre symbol.

On the basis of my computation, I pose the following conjecture.

Conjecture. Let $$p$$ be an odd prime, and let $$c,d\in\mathbb Z$$. Suppose that $$F_p^{(n)}(c,d)=0$$, where $$n$$ is a positive integer.

(i) If $$p\equiv1\pmod4$$, then $$\left(\frac{c^2-4d}p\right)=1\not=\left(\frac dp\right).$$

(ii) If $$p\equiv3\pmod 8$$ and $$p\nmid d$$, then $$\left(\frac{c^2-4d}p\right)=-1.$$

I think this conjecture is at the research level. Any ideas towards the solution?

• One part of your conjecture is rather easy. Assume $p\equiv1\pmod4$. In that case your sum has an odd number of terms, each either $\pm1$ or $0$ (assuming you declare the Legendre character to vanish at zero). So unless at least one of the terms is zero, the sum is odd. That is a contradiction. Therefore the polynomial $x^{4n}+cx^{2n}+d$ has a zero in the field $\Bbb{F}_p$. Implying that the quadratic $x^2+cx+d$ also has a zero in the prime field. This means that the discriminant of that quadratic must be a square. – Jyrki Lahtonen Dec 28 '18 at 14:17