$Ax^2 + By^3$ representing infinitely many primes

Question

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?

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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.

Answer 1

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 $|n| < 10^{4}$.

Answer 2

Let's focus on $A = B = 1$. Although $X^2+Y^3$ is absolutely irreducible, it is possible to find some structure of solutions to $X^2+Y^3 = mn$ required by bilinear forms arising in approaches to the expected number of prime numbers of a given form.

Indeed, let $H$ denote a positive definite 2 x 2 symmetric matrix with integer entries and let $v$, $w$ denote 2-dimensional column vectors with integer entries. Then for:

$m:= v^T Hv$, $n:= w^T Hw$, $x: = v^T Hw$, $y: = \det[v|w]$

the following identity holds:

$$mn = x^2+\det(H)\cdot y^2.$$

Thus, if we impose the condition:

$(*)$ $$\det(H)=\det[v|w],$$

we get factorisations $x^2+y^3=mn$. However the condition $(*)$ is unwieldy, at least in part due to the stipulation that $H$ is symmetric.

Suppose we are counting prime numbers below some threshold $N$. Then, roughly, entries of $H, v, w$ are resp. ~$N^{1/6}$, ~$N^{\alpha}$, ~$N^{1/3-\alpha}$ for $0<\alpha<1/3$. Probably some subinterval of $\alpha$ would be enough.

One may hope that after some number of squarings of the bilinear form, we will arrive at a feasible counting task, however it is not clear how to execute this.

An idea is to decompose $H = L^T L$ for 3 x 2 integral matrices $L$ whose entries are ~$N^{1/12}$. This step should be worked out, because the correspondence $H$ vs $L$ doesn't hold exactly. One advantage we obtain here is symmetry of $H$ for all $L$. Moreover, squarings of the bilinear form introduce the conditions such as $m_1 = m_2$, which means $v_1^T H_1 v_1=v_2^T H_2 v_2$, so $||L_1 v_1||^2=||L_2 v_2||^2$. This can be guaranteed if we are not economical when squaring, by stipulating that $L_1 v_1 =L_2 v_2 \in\mathbb{Z}^3$. This is the second advantage which looks promising since now we have linear equations instead of quadratic, nevertheless it works only for $\alpha$ near the middle, ie. $1/6$. A worse thing is we still have $(*)$. Using cross product for the columns of $L$, we can see that we would have a chance of succeeding, given an asymptotics for the following counting task (which corresponds to $\alpha=1/6$) where $k$ is some sufficiently large power of 2:

$$||u_i||^2=\pm\det[u_{i-1}|u_i|u_{i+1}]/d_{i}d_{i+1},$$

for $i = 1, 2, ..., k$ (cyclic indices) where $u_i$ are 3-dimensional column vectors with integral entries ~$R$ and $d_i:=\gcd(u_{i-1}\times u_i)$ are of the magnitude $R^{1/2}$. So, the asymptotics should be of the order approximately $R^{k/2}$.