I want to prove the following:
$$\forall x\in\mathbb{N}\ \text{ there always exists a prime }p\equiv1 \pmod 6 \text{ s.t. }p|(2x)^2+3;$$$\ \text{i.e. } (2x)^2\equiv-3 \pmod p$ where $p$ is some prime sufficing $p\equiv1 \pmod 6$ has to be true for all $x\in\mathbb{N}$.
I want to represent $2x$ as a product of 2 and primes $p_i$ of the form $6k_i+1$; i.e. $p_i\equiv1(\mod 6)$ and $2x=2\prod_{i=1}^{n}p_i$. Note that $x\in\mathbb{N}$ is chosen arbitrarily beforehand. Is this even possible?
What I've tried so far: I've prove with the Legendre symbol that $\big(\frac{-3}{p}\big)=1$ when $p\equiv1(\mod6)$ which clarifies the existence of an element $y\in\mathbb{Z}/(p)$ s.t. $y^2\equiv-3(\mod p)$. But since I want to show that it holds for all $y=2x$ where $x$ is chosen, this is useless i.m.o..
So maybe brainless trial and error will get me somewhere:
$x=1$: $2^2+3\equiv0(\mod p)$ where we choose $p=7\equiv1(\mod6)$ prime.
$x=2$: $4^2+3=19\equiv0(\mod p)$ where we choose $p=19\equiv1(\mod6)$ prime.
...
$x=a$: $4a^2+3\equiv0(\mod p)$ for some $p\equiv1(\mod6)$ iff (?) $4a^2+3\equiv1(\mod6)$.
This seems a dead end.
I also was considering $\big(\frac{-3\cdot4^{-1}}{p}\big)$, but this is difficult to compute since we don't know the inverse of 4 explicitly modulo $p$. Of course there's an algorithm for it, but I doubt it will bring succes in computing the Legendre symbol.
Still, $x\in\mathbb{N}$ is fixed, hence we cannot just manipulate $x$. Clearly, I'm stuck. Could anyone get me on the right track? All help is appreciated!
EDIT:
First title that is now deleted: For every $x\in\mathbb{N}$ write $2x=2\prod_{i=1}^{n}p_i$ with $p_i=6k_i+1$ primes