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Question: Consider the set $$ A(m) : = \{ (a, b) \in \mathbb{N}^2 : \; N \leq a \leq 2 N, \; a^2 - b^2 = m \}, $$ where $m \in \mathbb{Z}$ and $N \in \mathbb{N} \setminus \{ 0 \}$. Then $$ \sup_{m \in \mathbb{Z} \setminus \{0\}} (\sharp A(m)) = O(N^{\epsilon}) \; \mbox{ for all } \epsilon > 0 \; ? $$

Remark: My failed tentative was to use the following result related to divisor bounds of a natural number: $$ \sharp B(\ell) : = \sharp \{ (c, d) \in \mathbb{Z}^2 : |c | \leq 2 N, |d| \leq 2 N \mbox{ and } c \cdot d = \ell \} = O(N^{\epsilon}),$$ but, in the above case we not have information about the localization of $b \in \mathbb N$. Thanks in advance !

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Case 1: $m>0$, then $m \leq 4N^2$, so $d(m)=O((4N^2)^{\epsilon})= O(N^\epsilon)$, but then $c=a-b, d=a+b$ are divisors of $m$ so the number of such pairs lies in a $O(N^\epsilon)^2=O(N^\epsilon)$ set, hence $2a=c+d$ is in a $O(N^\epsilon)$ set

Case 2: $m<0, n=-m$ and assume there are at least two distinct solutions $(a,b), (a_1,b_1), N \leq a < a_1 \leq 2N, b^2-a^2=b_1^2-a_1^2=n>0$ as otherwise there is nothing to prove.

Then since $b_1 > b$ it follows that $3N^2 \geq a_1^2-a^2 =b_1^2 - b^2 \geq 2b+1$, so $b=O(N^2), n=|m|=O(N^4), d(m)=O(N^\epsilon)$ and the method from case 1 applies so we are done

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  • $\begingroup$ Dear @Conrad, Case 1 was clear to understand, because $m > 0$ implies $b \in [0, 2N]$ say. But Case 2 remains unclear. $\endgroup$
    – Marcelo Ng
    Commented Mar 18, 2019 at 22:17
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Commented Mar 18, 2019 at 22:23
  • $\begingroup$ 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 $b$ ? $\endgroup$
    – Marcelo Ng
    Commented Mar 18, 2019 at 23:54
  • $\begingroup$ $2b+1 \leq b_1^2-b^2=a_1^2-a^2 \leq 3N^2$, so $b$ is bounded by ~$N^2$, hence $m$ is bounded by ~$N^4$ $\endgroup$
    – Conrad
    Commented Mar 19, 2019 at 0:05
  • $\begingroup$ Yes, but the same is not valid for $b_1$, so there exists a problem. $\endgroup$
    – Marcelo Ng
    Commented Mar 19, 2019 at 0:16

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