Are there any known results of the form

there are infinitely many primes of the form $Ax^2 + By^3$

for integers $A$, $B$?

Assuming there are currently no known results of this form, what is the outlook on such problems? That is, are these problems we could plausibly attack using known methods, or do they seem to be hopelessly out of reach for now?

Commentary: I'm aware of similar results of Heath-Brown on primes of the form $x^3 + 2y^3$ and of Friedlander and Iwaniec of primes on the form $x^2 + y^4$.

  • On one hand, the density of numbers of the form $Ax^2 + By^3$ is larger than those of the form $x^3 + 2y^3$ or $x^2 + y^4$ ($N^{5/6}$ versus $N^{2/3}$ or $N^{3/4}$). I would naively expect this to make it easier to prove that there are infinitely many primes of the former form.

  • On the other hand, the proofs by Heath-Brown and Friedlander and Iwaniec apparently rely on the factorizations of the polynomials $x^3 + 2y^3$ and $x^2 + y^4$ over extensions of $\mathbb{Q}$, and $Ax^2 + By^3$ does not have such a factorization.

So it's not clear to me what I should guess about the difficulty of this problem.

  • 6
    $\begingroup$ Both the Friedlander--Iwaniec and Heath--Brown theorems use that $x^3+2y^3$ and $x^2+y^4$ are specializations of norm forms in number fields. This is lacking in the $x^2 +y^3$ situation, unless one studies it in the more special case of $x^2 + y^6$. But in that case, one does not have enough control over levels of distribution to make the Friedlander--Iwaniec argument work (at least as things stand now). The $x^2 +y^3$ problem is of interest to researchers wanting to produce elliptic curves with prime conductor. People have thought about it, so far unscuccesfully. $\endgroup$ – Lucia Feb 4 at 22:08
  • $\begingroup$ Lack of factorization over number fields surely is going to be an issue, but I have found one reference which deals with the case $x^2+y^2+1$, which is absolutely irreducible: matwbn.icm.edu.pl/ksiazki/aa/aa16/aa1642.pdf $\endgroup$ – Wojowu Feb 4 at 22:10
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    $\begingroup$ The density of the sequences is the most crucial aspect for the congruential sums (Type I in the parlance), but not so much so for the bilinear sums (Type II) (though there is /some/ necessary density, by Cauchy-Schwarz considerations). Indeed, in the results cited, the arithmetic of the sequence plays the larger role in bounding the bilinear sums. On the other hand, there are examples where the bilinear estimates can come about from more flexible means, e.g. as in Duke-Friedlander-Iwaniec for equidistribution of quadratic congruences modulo primes (cf Chapter 21 of Iwaniec-Kowalski). $\endgroup$ – literature-searcher Feb 4 at 22:25
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    $\begingroup$ @Wojowu Motohashi's result is pretty much subsumed by Iwaniec's 1974 work (which is an application of the half-linear sieve IIRC). mathoverflow.net/questions/55384/… matwbn.icm.edu.pl/ksiazki/aa/aa24/aa2451.pdf $\endgroup$ – literature-searcher Feb 4 at 22:31
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    $\begingroup$ I agree with what Lucia says (that the only currently plausible attack on $x^2+y^3$ is via $x^2+y^6$ and the latter is (just) not dense enough), but also want to note that the handling of the bilinear terms in the two papers is quite different, with H-B's being a multi-dimensional large sieve (involving the cubic field), while F-I exploits the regularity of the squares. The fact that H-B can handle density 2/3 and F-I (just) cannot derives from the use of a Brun-type sieve over a short logarithmic range in the former (splicing type I and II in the other ranges), which is unavailable for F-I. $\endgroup$ – literature-searcher Feb 4 at 22:56

FYI, here's a table of numbers of such primes less than $10^{k}$:

|N       |number of primes |number of primes of the form x^2+y^3 |proportion|
|10      |4                |2                                    |0.5       |
|100     |25               |8                                    |0.32      |
|1000    |168              |40                                   |0.238095  |
|10000   |1229             |192                                  |0.156225  |
|100000  |9592             |1094                                 |0.114053  |
|1000000 |78498            |6098                                 |0.0776835 |
|10000000|665025           |35733                                |0.0537318 |

As Elkies said, this result only allows $x, y>0$. If we allow negative $x$ and $y$, then I found an interesting result: every prime $<10^{6}$ can be expressed as $x^{2} + y^{3}$ for some $x, y\in \mathbb{Z}$. However, I don't have any idea to prove this!

EDIT: I found that the above claim is wrong, and there was a problem of overflow. In fact, $p$ is a sum of square and a cube if and only if the elliptic curve $E_{p}:y^{2} = x^{3} +p$ has an integral point. Such an elliptic curve is called Mordell's curve, and here's OEIS link for $n$ such that $E_{n}$ has no integer points. As you can see in the list, $E_{7}$ has no integral solution, i.e. $7 = x^{3} +y^{2}$ has no integer solution. I think this paper might help - which find all the integral points for $|n| < 10^{4}$.

  • $\begingroup$ do you have a reference available for this result ? $\endgroup$ – Konstantinos Kanakoglou Feb 5 at 0:15
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    $\begingroup$ It's easier to compute such data than to find it in the literature . . . (assuming, as seems to be the case here, that only positive $x,y$ are allowed). $\endgroup$ – Noam D. Elkies Feb 5 at 0:26
  • $\begingroup$ You are probably right. I was just wondering whether similar sequences were recorded in OEIS $\endgroup$ – Konstantinos Kanakoglou Feb 5 at 0:38
  • $\begingroup$ @KonstantinosKanakoglou I just wrote a program with C++, and as Elkies said, I assume that both $x$ and $y$ are positive. I'll add a table that also allows negative $y$. $\endgroup$ – Seewoo Lee Feb 5 at 1:01
  • $\begingroup$ Careful -- that's much harder because of the possibility of massive cancellation. For example, the prime 971 occurs but not before y = -265. (There are much worse examples, but with smaller alternative solutions; e.g. 17 = 9+8 is also 378661^2 - 5234^3.) $\endgroup$ – Noam D. Elkies Feb 5 at 2:09

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