When is $y=3x^2+3x+1$ a prime number in $\mathbb{Z}$ with $x \in \mathbb{Z}$? The first few values of $y=3x^2+3x+1$ for integer values of $x$ are $7, 19, 37, 61, 91$, and $127$.  I am wondering under what conditions of $x$ is $y$ a prime number?    
I had initially hoped that Vieta's formula would produce something notable but was unsuccessful.  I believe that knowing $3x^2+3x+1$ factors as $\frac1{12}(-6ix + \sqrt3-3i)(6ix+\sqrt3+3i)$ could be useful, although I have not been able to make any further progress and would appreciate some help.  
I also wonder how the results on this might generalize over to other irreducible polynomials $ax^2+ax+1$, although I am still trying to pick apart the case for $a=3$.
 A: Obviously, this is not an very easy problem to tackle. I predict that there exists infinitely many $x$ for which $y$ is a prime number. Indeed, this problem seems to be a corollary of generalization of Dirichlet's Theorem on Arithmetic Progression which is still an open problem. We even don't know, for what $x\in\mathbb{Z}$, $x^2+1$ is a prime number.
A: Polynomial remainder theorem( more of a statement on binomial powers) will tell you that:


*

*$x=1$ producing 7, will knock out all $x\equiv 1\bmod 7$

*$x=2$ producing 19, will knock out all $x\equiv 2\bmod 19$
$\vdots$

*$x=z$ producing a number divisible by prime p, will knock out all $x\equiv z\bmod p$
What makes this polynomial have so many primes ( though $91=7\cdot 13$) to start, is that all entries must be odd, remainder 1 on division by 3, and can't ever be divisible by 5 either. So it makes small factors less likely. Just with polynomial remainder theorem knowledge, I did a sieve of $x$ values, in the first 127 values 51 succumb to factorization based on these 6 first terms. 
[5, 8, 12, 15, 18, 19, 21, 22, 26, 29, 31, 33, 36, 40, 43, 44, 47, 50, 54, 57, 59, 61, 64, 65, 68, 70, 71, 75, 77, 78, 82, 83, 85, 89, 92, 96, 97, 99, 103, 106, 109, 110, 113, 114, 116, 117, 120, 122, 124, 126, 127]
All these $x$ values need not be checked ( except maybe for more prime factors to eliminate with.).  We can bring in Fermat's little theorem, but that mostly tells us that cubes without 7 as a factor of the base, turn into $3x+4\bmod 7$.  Seems like overkill in this small range though, only cubes we haven't knocked out are 27 and 125, which both turn to remainder 1 mod 7.  $3x+4\bmod 11$ works on fifth powers like 32 the one under 243. But these both give back non zero residues of 1 and 7 ( the two possible for coprime powers mod any prime in this case, insert Legendre symbol.). 
These results using Fermat, can be generalized, Both the theorem, and the result $ax+(1+a) \bmod p$ for powers $x$ with  $p-1\over 2$ exponent. 
Addendum $$3x^2+3x+1=(2x+1)(x+1)+x^2$$  Now assume: $$2x+1\notin\mathbb{P}$$ by Sieve of Sundaram, we have $x=2ab+a+b$  which the makes it $b^2\bmod 2b+1$ . It's also, $-x\bmod 3x+1$ Just some polynomial division results. 
