4
$\begingroup$

Let's define an infinite sequence of positive integers as :

$a_n=k^2+(2n-1)k+2n-1 $ , where $ k,n \in \mathbf{Z^{+}}$

Suppose that one can prove that this sequence contains infinitely many prime numbers for any particular $k$. The first term of the sequence for any $k$ is of the form :

$a_1=k^2+k+1$

My question is : what is sufficient condition to prove that this polynomial produces primes for infinitely many values of $k$ ?

Note that there is strong experimental evidence that $k^2+k+1$ is prime for infinitely many values of $k$ and that this polynomial satisfies conditions of Bunyakovsky's conjecture.

  • 3
    I assumed that you knew that *no* quadratic polynomial is known that represents infinitely many primes. So I thought the question could not be about fixed $n$. The primality of $k^2+k+1$ for infinitely many $k$ is a well-known open problem. I am therefore deleting the answer, which dealt with fixed $k$ and variable $n$.2011-11-05
  • 0
    @Andre,I know that it isn't known whether such polynomial exist, that's why I have involved this sequence in my question..2011-11-05
  • 3
    Supposing isn't necessary; Andre answered that fixing $k$ allows infinitely primes in $a_n$ due to Dirichlet's Theorem (his answer is now deleted). Now would you mind explaining what $n$ and $a_n$ have to do with anything if your question is about $k^2+k+1$? Why the irrelevant, misleading build-up?2011-11-05
  • 0
    @anon,I think that fact that $k^2+k+1$ is the first term of sequence above may be interpreted as necessary condition2011-11-05
  • 2
    @pedja: I see, you are hoping that the additional parameter can help in settling the $k^2+k+1$ problem. Unfortunately each value of $n$ leads to another open problem. The fact that for each $k$, there are infinitely many $n$ for which we get a prime gives no information about $n=1$, or any other value of $n$. Each "row" can have infinitely many primes, while each "column" has only finitely many. It is not hard to come up with examples of such situations, though I certainly cannot for your two-parameter family.2011-11-05
  • 0
    @Andre,Do you think that we may claim that there exist at least one "column" with infinitely many primes ?2011-11-05
  • 0
    @pedja: My earlier comment was to the effect that in *general* situations, knowing that each row has infinitely many primes does not show that any column does. But one cannot rule out the possibility that for *this family* one could push such an argument through. I am not optimistic, but would love to be wrong.2011-11-05
  • 2
    I think if we could make such a claim, someone would have done so some time in the last 200 years. As no one has....2011-11-05
  • 0
    @Andre,thanks for your comments..2011-11-05
  • 0
    Related: https://math.stackexchange.com/questions/12849632018-11-30

1 Answers 1

1

I don't think there is any comprehensive conjecture on the distribution of prime values of polynomials in two or more variables. Other than simply being more general, when there are multiple variables there are algebraic curves along which the density of primes is higher or lower. This is analogous to algebraic surfaces in which it is possible for all the rational points to lie on a lower-dimensional subset such as a line, or for the surface to have many rational points but some curves on the surface will have no rational points at all.

Iwaniec and Friedlander proved that there are infinitely primes of the form $x^2 + y^4$ but there are infinitely many values of $x$ for which the polynomial is always composite. Any asymptotic prediction of the distribution of prime values in the sequence must have some non-uniformity when expressed in terms of $x$ and $y$, and there is no known way to "concentrate" the theorem down to statements about infinitely many primes in subsequences with $x$ (or $y$) constant.