Timeline for Upper bound for a subset of $\mathbb{N}^2$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2019 at 1:36 | comment | added | Conrad | So, the set $A(m)$ above having at least two elements imposes a fairly strong constraint on the size of $|m|$, namely being bounded by ~$N^4$; intuitively this is actually clear (not the precise bound, but that a power bound must exist) since if $|m|$ is much bigger than $a$ it follows that $b$ is about $\sqrt{|m|}$ and $b-a, b+a$ are both about $b$ as $a$ is negligible, but there is no way a number has two distinct pairs of complementary divisors ($d, \frac{|m|}{d}$) this close to its square root for a sufficiently small "this close" and the computations above put a numeric value on that | |
| Mar 19, 2019 at 0:54 | comment | added | Conrad | It doesn't matter since the bound for $|m|$ matters - note that actually it implies bound for $b_1$ too since $b_1^2 \leq |m|+ 4N^2$, so $b_1$ is bounded by ~$N^2$ | |
| Mar 19, 2019 at 0:16 | comment | added | Marcelo Ng | Yes, but the same is not valid for $b_1$, so there exists a problem. | |
| Mar 19, 2019 at 0:05 | comment | added | Conrad | $2b+1 \leq b_1^2-b^2=a_1^2-a^2 \leq 3N^2$, so $b$ is bounded by ~$N^2$, hence $m$ is bounded by ~$N^4$ | |
| Mar 18, 2019 at 23:54 | comment | added | Marcelo Ng | I realize that $b_1 > b$ implies $b_1 \geq b + 1$ . Therefore,$b_1^2 \geq (b + 1)^2$ and $b_1^2 - b^2 \geq 2 b + 1$. But, the existence of two solutions no implies necessarily a global bound for the value of $b$ ? | |
| Mar 18, 2019 at 22:23 | comment | added | Conrad | Why? I showed that if there are at least two solutions $|m| = O(N^4)$, so the same reasoning applies, number of divisors of $m$ is $O(N^{\epsilon})$, pairs of such same, $a$ is the semisum of such, so same bound applies | |
| Mar 18, 2019 at 22:17 | comment | added | Marcelo Ng | Dear @Conrad, Case 1 was clear to understand, because $m > 0$ implies $b \in [0, 2N]$ say. But Case 2 remains unclear. | |
| Mar 18, 2019 at 21:58 | history | answered | Conrad | CC BY-SA 4.0 |