**Update**:

The chat room for this question: http://chat.stackexchange.com/rooms/69953/discussion-between-fedja-and-abiessu

The problem statement remains unchanged (below).

Consider the following problem:

Given the following sets with $u \in\Bbb Z^+$: $$\begin{align}A_u&=\{x^2:x\in [2^{u-1},2^u-1],\exists s,t \in\Bbb Z^+ : x^2=s^3+2s^2+st+t\}\\B_u&=\{x^2:x\in [2^{u-1},2^u-1],\exists s,t \in\Bbb Z^+ : x^2=2s^3+2s^2+2st+t\}\end{align}$$

prove that $\exists N\in\Bbb Z^+$ such that $\forall u\gt N, |A_u|\ge|B_u|$.

**Edit**: it has been pointed out that this problem directly correlates with the Twin Primes conjecture. This does not change the question.

My approach is to rearrange the specifier equations as follows (using $s_A,t_A,s_B,t_B$ to differentiate the sets):

$$x^2=(s_A+1)(s_A^2+t_A)+s_A^2\\ y^2=(2s_B+1)(s_B^2+t_B)+s_B^2$$

From here, it is almost "obvious" that the problem statement should be correct, and I take the path of letting $s_A=2s_B$ for all but $s_A=1$ and comparing counts of values for each such pairing. Once I have counted all the differences where $s_A=2s_B$, then I go back and count all the overlaps between $s_A=1$ and $s_A\gt 1$. When all of this is done, I get a value $N=45$ (which I am certain could be improved).

Is there a more effective or efficient approach? With such an "obvious" problem statement, it seems like there should be an easier way to get the required results...

Addendum:

I glossed over the details above, but the counting of actual results goes like this: for each value of $s_A=2s_B$ where $s_A+1$ is prime, there are exactly two possible solutions of the congruence $s_A^2\equiv x^2\pmod{s_A+1}$, and there are exactly two possible solutions of $s_B^2\equiv x^2\pmod{2s_B+1}$. These solutions exist for both congruences. Therefore the arithmetic sequences in $t_A,t_B$ given by $(s_A+1)t_A+(s_A+2)s_A^2$ and $(2s_B+1)t_B+(2s_B+2)s_B^2$ each produce the same number of values $x^2,y^2$ within a given interval whenever $s_A+1=2s_B+1$, up to a maximum difference of two values produced (per prime value $s_A+1$). The squarefree non-prime values of $s_A+1$ account for double-counted values, and if we account all the "maximum difference of two" possibilities in favor of $B_u$, we should effectively count the number of values that $s_A\gt 1$ can take on which affect the given interval and multiply it by $2$ as the "worst case" for the value of $|B_u|$. For the overlap counting between $s_A=1$ and $s_A\gt1$, we account for the "worst case" by taking the fact that $s_A+1=2$ covers all odd squares within any interval for $u\gt 5$, then multiply this result by all the overlap possibilities for each prime greater than $2$ up to the maximum possible value of $s_A+1$ as $\left(1-\frac 23\right)\left(1-\frac 25\right)\dots\left(1-\frac 2p\right)$, at which point we apply the result

$$\left(\prod_{p=3}^n\left(1-\frac 2p\right)\right)^{-1}=\frac 14e^{2\gamma}\Pi_2^{-1}\log^2n+O\left(e^{-c\sqrt{\log n}}\right)$$

(from https://math.stackexchange.com/a/22435/86846).

This question is cross-posted from Math.SE (https://math.stackexchange.com/q/2521575/86846) following an intense period of no activity whatsoever.

reallyhard even to somebody who cannot recognize it or has never heard of it. Normally I wouldn't spoil the fun, but attaching a bounty to it is a bit too much to my taste :lol: $\endgroup$ – fedja Dec 8 '17 at 21:09didknow that your question implied the twin prime conjecture. If you decide to rewrite your opus in a way more aligned with the standards of mathematical writing, I promise to give it one more try. As of now, I'd rather have my evening meal :-) $\endgroup$ – fedja Dec 9 '17 at 0:3922more comments