Can the proof of one conjecture be considered a proof of the other conjecture?

The general method of building an infinite number of prime producing quadratic polynomials was given in the link below.

What can we learn from prime generating polynomials?

Here we will give just one example of two prime producing polynomials which can generate twin primes. We start with the following polynomial that can only produce even integers and we add a "prime seed part" to it to make it generate primes.

e(n) = (10+n)(1+n) = n^2 + 11n + 10

then we add the "prime seed part" 2n+1 to it to get:

p(n) = n^2 + 13n + 11

Now to get a twin prime we need another prime producing polynomial, g(n), which differs by 2 from p(n).

g(n) = p(n) + 2 = n^2 + 13n + 13

For n=0,...,10 the combination of these two polynomials produced 5 pairs of twin primes.

So does the proof of one conjecture lead to the proof of the other conjecture?


2 Answers 2


I don't know a proof for the implication you stated when we are talking about the simple Bunyakovsky conjecture. ( It seems in the beginning that the implication is true by construction of two polynomials $P$ and $Q$ such that $P-Q=2$ and apply the conjecture but it's not obvious like i explained below)

The simple Bunyakovsky conjecture states that: any irreducible polynomial with the positive leading coefficient such that :as $n$ runs over the positive integers, the numbers $f(n)$ should be relatively prime, then $f(n)$ is prime for infinitely many positive integers $n$.

Now as you stated we can find two polynomials $P$ and $Q$ which satisfy the conditions of Bunyakovsky conjecture and $P-Q=2$, so $P(n)$ and $Q(n)$ are prime for an infinite number of value of $n$. But in order to prove the twin prime conjecture we need $P(n)$ and $Q(n)$ to be simultaneously prime.

The generalized Bunyakovsky conjecture which was in fact the original, it states that given $n$ polynomial satisfying the conditions above and for any prime $p$ the values of these polynomials are coprime with $p$ for some integer, then these polynomial produces simultaneously an infinite number of primes, this conjecture implies a lot of others:

There is more general conjecture called Schinzel's hypothesis H.

  • $\begingroup$ Elaqqad, could you please be more specific with what you called "there is problem in the obvious thought you have". I am here to learn. I thought about it but I can't figure out what you meant. $\endgroup$
    – user25406
    Commented Apr 7, 2015 at 20:19
  • $\begingroup$ I'm sorry if what I wrote is not what you thought! $\endgroup$
    – Elaqqad
    Commented Apr 7, 2015 at 20:27
  • $\begingroup$ I am fine with what you wrote and there was really no need to edit your post. I thought you were pointing to something specific that is wrong. The big problem is, like you pointed it out, that there is no reason for these two polynomials to produce simultaneously an infinite number of primes that differ by 2 only though there is no reason for them not to produce an infinite number of primes. So now we run into another problem, can these two conjectures be true yet not imply each other? $\endgroup$
    – user25406
    Commented Apr 7, 2015 at 20:36

You ask:

Can the proof of one conjecture be considered a proof of the other conjecture?

Results of the form "well-known conjecture $\varphi$ implies well-known conjecture $\psi$" are indeed interesting and important. For example, Wiles didn't directly prove Fermat's Last Theorem; what he actually proved was the modularity theorem for semistable elliptic curves. Since this was already known to imply Fermat's Last Theorem, it was enough.


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