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Let $p_m$ be the $m$th positive prime number in $\Bbb{Z}$. Then $f \in \Bbb{Z}[X]$ is irreducible if:

$$ \liminf\limits_{m \to \infty} \dfrac{\# \{f(n) \text{ is prime } : n \lt p_m \}}{m} \gt 0 $$

I think I have an elementary proof for it.

Proof: If the polynomial $f$ is reducible, then $f(n) = a(n)b(n)$ and so $f(n)$ can only be prime a finite number of times, no matter how far out $m$ goes. Insert argument about how a polynomial (namely $a$ or $b$) cannot infinitely take on values $\pm 1$, maybe talking about Taylor series for $\cos$ since the approximating polynomials "try" to do this.

Under an asymptotic density limit, if the top is finite, then clearly the $\lim \to 0$. $\square$

What do you think? How would you prove the converse (probably way harder)? Have I solved twin primes? J/k :)

EDIT:

Changed $\lim $ to $\liminf$ since who knows how the graph of the counting function behaves.

EffingLoveMath
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    There are irreducible polynomials that never take on prime values, for example because they are automatically even. And almost nothing is known in general. For example it is not even known whether $x^2+1$ is prime infinitely often. But there are reasonable conjectures. – André Nicolas Mar 31 '16 at 01:02
  • @AndréNicolas oh, so the converse is not true. Additionally, such reverse-of-the-above statements would be extremely hard to prove. I see. – EffingLoveMath Mar 31 '16 at 01:03
  • @AndréNicolas by automatically even do you mean $2f(X)$? – EffingLoveMath Mar 31 '16 at 01:19
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    That's kind of boring. More like $x^2+x+8$. One can replicate the same phenomenon with higher moduli. – André Nicolas Mar 31 '16 at 01:31
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    If I may slightly expand on Andre's idea, for a fixed prime $p$, the polynomial $x^p-x+3p$ is divisible by $p$ for every $x$ and is never prime. – Arkady Mar 31 '16 at 01:43

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