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 $R^{k/2}$.

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