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?

  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 28 '18 at 14:17

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