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As far as I know, we still do not have a proof that some second degree polynomial in one variable with integer coefficients takes an infinite number of prime numbers as its values, even the "simplest" ones like $f(n)=n^2+1$ are still out of reach, to say it that way.

But, for the purpose of this question we will have to assume that there are polynomials of second degree in one variable with integer coefficients that take an infinite number of prime numbers as its values, so the answer to this question must depend on some unproven facts.

Let us now denote by $P_{2}$ the set of all second degree polynomials with integer coefficients such that every member of the set takes an infinite number of prime numbers as its vales and suppose that the set $P_{2}$ has at least two elements.

Let us now choose two polynomials $a$ and $b$ from the set $P_{2}$ and denote by $p_a(n)$ the number of prime numbers in the set $\{a(1),a(2),...,a(n)\}$ and denote by $p_b(n)$ the number of prime numbers in the set $\{b(1),b(2),...b(n)\}$.

The question is:

Is it true that for every $a$ and $b$ from the set $P_{2}$ we have $\lim_{n \to \infty} \dfrac {p_a(n)}{p_b(n)}=1$?

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Interestingly, no! The Bateman-Horn conjecture gives a widely-believed asymptotic formula for $p_a(n)$ and $p_b(n)$. In particular, this conjecture implies that $$ \lim_{n\to\infty} \frac{p_a(n)}{p_b(n)} = \prod_p \frac{p-N_a(p)}{p-N_b(p)}, $$ where the product is over all primes and $N_a(p)$ is the number of integers $m\in\{1,\dots,p\}$ such that $p$ divides $a(m)$ (and similarly for $N_b(p)$). There's no reason this infinite product should equal $1$, and indeed it can be evaluated for your favorite polynomials $a$ and $b$. For example, if $a(x)=x^2+1$ and $b(x)=x^2+3$, then the ratio equals $\sqrt{\frac32}$.

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  • $\begingroup$ Indeed an interesting conjecture. $\endgroup$ – Farewell Apr 22 '16 at 2:34

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