# 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

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).

• Your title says $3x^2 + 3x + 1 = 2p - 1$ but your question text says $3x^2 + 3x + 1 = 3p - 2$. Which one, if either, is it? – John Omielan Jul 23 at 2:06
• There seems to be no obvious connection between these sets of primes: OEIS sequence A002383 for $3x^2+3x+1=3p-2$, i.e. $p=x^2+x+1$ and A002407 for $3 y^2 + 3y + 1$. Of course the Bunyakovsky conjecture would imply both sets are infinite. – Robert Israel Jul 23 at 2:13
• @JohnOmielan fixed. – heepo Jul 23 at 2:24
• @RobertIsrael I mean to prove this under my assumption above as a stronger alternative to assuming the Bunyakovsky conjecture. – heepo Jul 23 at 2:27
• @heepo I agree that this implication cannot be shown. – Peter Jul 23 at 7:41

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.

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.