If there exist infinitely many $x \in \mathbb{Z}:3x^2+3x+1 = 3p-2$ for $p \in \mathbb{P}$, show there exist infinitely many $y:3y^2+3y+1$ is prime

Question

Assume there exist infinitely many $x$ such that: $$3x^2+3x+1 = 3p-2$$ Where $p$ is prime. Can it be shown there exist infinitely many $y$ such that: $$3y^2+3y+1=q$$ Where $q$ is prime? I believe that it cannot be shown as our assumption tells us nothing of which primes exist such that $3p-2 = 3x^2+3x+1$ and so knowing there exist infinitely many primes of the form $3p-2$ does not help us, but maybe I am wrong (perhaps it can be shown I am wrong with a relevant proof).

Answer 1

Here's my take, if, $q$ replaces $p$ we get by:$$3p-2=3(p-1)+1$$ that, $$9y^2+9y+1=3x^2+3x+1\implies 3y^2+3y=x^2+x$$ which has solutions: $$(x,y)\in\{(2,1),(9,5),(35,20)\}$$ less than $(100,100)$

possible values of $p-1$ are:

0,2,6,12,20,30,42,...

because $p-1$ must be the product of consecutive integers.

If the $q$ replacing $p$ case works then $9y+10\equiv 3x+4 \bmod r$ for any value pairs $x,y$ both congruent to powers with exponent $r-1\over 2$ mod prime $r$. We also get $$3y+1\equiv x+1\bmod r$$ via the $$ax^2+ax+c\equiv ax+(c+a)$$ reduction with constant term $c=0$. The first reduction, gives us $$x\equiv 3y+2\bmod r\implies x-1\equiv 3y+1\bmod r$$ contradicting the second coming from the same algebra.

Answer 2

For your first equation, simply adding $2$ to both sides yields $$x^2+x+1=p$$ so it suffices to prove that there exists two integers $a_x$ and $b_x$ such that $a_xb_x=1-p$ and $a_x+b_x=1$. $$\therefore p=a_x+b_x-a_xb_x$$

Same for how $$y^2+y+1=\frac{q+2}{3}$$ where $q=a_y+b_y-3a_yb_y$.

Clearly $ab\neq 0$ so if $a+b=1$ then one of $a$ or $b$ is negative. Now since all primes $p>2$ are odd, let $a$ and $b$ both be odd, or opposite parity. Then note that if $p>3$ then $p\equiv \pm 1\pmod 6$ and it should pretty much be trivial hereafter.