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Is there a non zero polynomial $R \in \mathbb{Z}[X,Y]$ such that there exists an infinite number of pair $(p,q)$ with $p$ and $q$ primes, $p \neq q$ and $R(p,q)=0$ ?

I know the curve must be of genus $0$ (Faltings-Mordell).

My question is related to Polynomial equations in $n$ and $\phi(n)$ that has been solved.

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Conjecturally, $R(x, y) = x - y + 2$ or $R(x, y) = x - 2y - 1$... –  Qiaochu Yuan Dec 14 '11 at 16:47
For the record, the conjectures mentioned in Qiaochu Yuan's comment are about twin primes and safe primes or Sophie Germain primes. –  lhf Dec 14 '11 at 16:54
Thanks for your answer. –  francis-jamet Dec 14 '11 at 17:11
Conjecturally, there are infinitely many primes $p$ such that $p^2-2$ is prime, so $R(x,y)=x^2-y-2$ will do. More generally, it is widely believed that if $f$ and $g$ are irreducible polynomials with integer coefficients, and if there is no $d\gt1$ such that $d$ divides $f(n)g(n)$ for all $n$, then there are infinitely many $n$ such that $f(n)$ and $g(n)$ are both prime. But the proofs are way out of reach. See Schinzel's Hypothesis H. –  Gerry Myerson Dec 14 '11 at 23:24
Less trivially, (p-q-2)(p-q-4)(p-q-6)(p-q-8)(p-q-10)(p-q-12)(p-q-14)(p-q-16)(p-q-18)(p-q-20) is zero infinitely often under Elliott-Halberstam. –  Charles Feb 9 '12 at 19:02

1 Answer 1

Meanwhile, we know the answer is yes. As of 2013 we know (thanks to Y. Zhang) that there is some $C\leq7\cdot10^7$ such that there are infinitely many prime pairs that differ by $C$. So we can take $R(x,y)=x-y-C$, or, if you want an explicit example, take $\prod_{\substack{C=2\\C\text{ even}}}^{7\cdot10^7}(x-y-C)$. As of april 2014 it is known that we can have $C\leq246$.

See here for more information.

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