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I am interested in primes and some of their varying characteristics which make them “qualitatively” different such as whether each prime generated by a certain type of Dirichlet arithmetic progressions belongs or not to the class of primes that are obtained from powerful order terms of that progression. Particularly, I wonder whether one can go beyond Dirichlet Theorem and establish that there is an infinite supply of this sort of primes in these progressions.

The case I am particularly interested is the case when d=1 and a=2k with k positive integer, but I would like to formulate the following Conjecture for the general case

For any two coprime positive integers a and d, there are infinite primes of the form $an^2m^3+d$ with n and m integers >=0

If true this would strengthen Dirichlet theorem on arithmetic progressions selecting only the terms of the progressions whose order is a powerful number (a number that if prime p divides it, $p^2$ also divides it).

I wonder about the likeliness of the conjecture (which intuitively seems high) and whether there is any proof I am still not aware of. I welcome estimates on the toughness of the proof, as well as any hints towards its development.

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    $\begingroup$ It is unlikely. The number of powerful integers less than $x$ is asymptotic to $\frac{\zeta(3/2)}{\zeta(3)}x^{\frac{1}{2}},$ which is the same order of magnitude as the counting function for the squares. Since the question of whether or not there are infinitely many primes of the form $qn^2+a$ is open and notoriously difficult, I doubt any progress has been made for powerful numbers. The simplest progression, the question of whether there are infinitely many primes of the form $n^2+1$, is known as Landau's conjecture, and has been open for 100 years. $\endgroup$
    – Eric Naslund
    Commented Jun 11, 2012 at 14:01
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    $\begingroup$ On the other hand, Friedlander and Iwaniec proved that there are infinitely many primes of the form $x^2+y^4$ and Heath-Brown proved that there infinitely many primes of the form $x^3+2y^3$, which are likewise sparse. So the question might be just on the border of doable. $\endgroup$
    – Felipe Voloch
    Commented Jun 11, 2012 at 14:23
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    $\begingroup$ @Felipe: Those are two hard papers, and the density of the smaller of those two sets is $\approx N^{2/3}$ in the interval up to $N$. Proving an infinitude of primes on a set of density $N^{1/2}$ would be a huge breakthrough. Perhaps it depends on what you mean by "on the border of doable." In a sense every problem is on the border of doable as you can never rule out the possibility of a new idea. $\endgroup$
    – Eric Naslund
    Commented Jun 11, 2012 at 20:58

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I'm sorry this isn't much of an answer, but you might be interested in this paper of De Koninck, Kátai and Subbarao. Your conjecture is almost strong enough to mean that for any squarefree $a$, there are infinitely many primes $p$ where the squarefree part of $p-d$ is exactly $a$ (the difference being that you do not require $(a,mn)=1$).

Section 4 in the cited paper establishes a sort of converse formulation: for any powerful $K$, there are infinitely many primes $p$ where the powerful part of $p-1$ is exactly $K$ (with the expected asymptotics). Of course this is a far denser set in which to look for primes, so this is certainly a much easier problem. Presumably no significant complications arise if we shift this by $d$ rather than $1$.

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    $\begingroup$ Erick i am very grateful for your answer and effort of summarizing section 4 of the paper. Its converse formulation pointed me a solution to the hidden question behind my conjecture. I was looking for infinite primes of a certain quality in Dirichlet Arithmetic progressions. They did not have to come from powerful order terms. Square free order terms is just fine for shifted primes with a powerful part K. From my reading of the paper the converse formulation is proved there up to page 529. Do you confirm my interpretation? Thank you very much. I owe you an aknowledgement in a future paper. $\endgroup$
    – Fernando Pimentel
    Commented Jun 15, 2012 at 16:17
  • $\begingroup$ @Fernando: That is truly nice to hear! I do agree with your interpretation, and I hope you are able to make good use of this result. I'll be interested to see your paper when it is ready. $\endgroup$
    – Erick Wong
    Commented Jun 16, 2012 at 5:42

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