Is $8x+5$ always square non-residue modulo $8x^2+7$ for natural $x$?

Question

Is $8x+5$ always square non-residue modulo $8x^2+7$ for natural $x$?

This holds for $x$ up to $10^8$. The kronecker symbol is never $1$ also.

Working modulo $8$ doesn't appear straightforward since $16x+5$ is often square modulo $8x^2+7$.

If necessary, assume they are coprime, $x \not \equiv 2 \pmod{3}$.

Answer 1

When working in $\Bbb Q(\sqrt{-14})$, one finds that $(-2)(5+2\sqrt{-14}) = (\sqrt{-14}-2)^2$, and so :

$(4x-2)^2 = -2(8x+5)$ modulo $8x^2+7$

So you only need to show that $-2$ is not a square.

Answer 2

You don't need to use advanced terminology of the other answerer, at least surely not for the case when $8x+5$, $8x^2+7$ are coprime (I haven't solved the other case).

As you've noticed, a common prime divisor of $8x+5$, $8x^2+7$ could only be $3$. Below is a proof in case anyone is interested.

If $8x+5$, $8x^2+7$ weren't coprime, then a prime $p$ would divide both.

$$(8x)x+7\equiv (-5)x+7\pmod{p}$$

$$\equiv -5x+7+8x+5\equiv 3x+12\equiv 0\pmod{p}$$

$$\implies x\equiv \frac{-12}{3}\equiv -4\pmod{p}$$

$$\implies 8x+5\equiv 8(-4)+5\equiv -27\equiv 0\pmod{p}$$

$$\implies p\mid 27\implies p=3$$

Therefore a common divisor of $8x+5$, $8x^2+7$ can only be a power of $3$.

$8x+5$, $8x^2+7$ are not coprime if and only if $8x+5\equiv 0\pmod{3}$ and $8x^2+7\equiv 0\pmod{3}$. If and only if $x\equiv 2\pmod{3}$.

If $8x+5$, $8x^2+7$ are coprime, then, as noticed by another answerer, $$(4x-2)^2\equiv -2(8x+5)\pmod{8x^2+7}$$

(because it's equivalent to $16x^2-16x+4\equiv -16x-10\pmod{8x^2+7}$, i.e. $2\left(8x^2+7\right)\equiv 0\pmod{8x^2+7}$, which is true)

$$\iff -2\equiv ((4x-2)(8x+5)^{-1})^2\pmod{8x^2+7},$$

so $-2$ is a square mod $8x^2+7$, contradiction, because $-2$ would be a square mod a prime divisor $p$ of $8x^2+7$ of the form either $8x+5$ or $8x+7$ (if all prime divisors of $8x^2+7$ were of the form $8k+3$ or $8k+1$, then $8x^2+7$ would itself be of the form $8k+3$ or $8k+1$, contradiction), contradiction, because $-2$ is a square mod an odd prime if and only if the odd prime is of the form $8k+1$ or $8k+3$ (see this link; it's related to quadratic reciprocity).