4

Let $H,M\geq 1$ and let $h_0$ and $m_0$ be fixed integers with $(h_0,m_0)\in [H,2H]\times[M,2M]$. Let $\alpha$ be a positive real number. I'm trying to find an upper found for the number of integer pairs $(h,m)\in [H,2H]\times[M,2M]$ such that $$ \tag{1} m h^\alpha = m_0 h_0^\alpha. $$ If $\alpha$ is rational, say $\alpha = \frac{p}{q}$, then $m^q h^p = m_0^q h_0^p$, and so the number of solutions is at most the number of divisors of $m^q h^p$, which is $O((HM)^\varepsilon)$ for any $\epsilon > 0$, the implied constant depending on both $\alpha$ and $\varepsilon$.

When $\alpha$ is irrational, I expect that the same bound should hold, but I'm not sure how to show this. Thus my question,

Is the number of solutions to (1) bounded by $O((HM)^\varepsilon)$ when $\alpha$ is irrational?

For context, the application I have in mind is a bound for certain exponential sums in the spirit of the classical paper "Exponential sums with monomials" by Fouvry and Iwaniec.

Any feedback and references are most appreciated.

  • If $\alpha$ is irrational, doesn't it imply $(m,h)=(m_0,h_0)$? – Zach Teitler Dec 06 '23 at 02:47
  • 1
    No, because $\alpha$ could be a quotient of logarithms, say $\alpha = \frac{\log(m/m_0)}{\log(h_0/h)}$. One then needs to show that this equation doesn't have too many solutions in $(h_0,m_0)$, which is essentially an equivalent problem. – Joshua Stucky Dec 06 '23 at 03:36
  • Given two solutions $(h_1, m_1), (h_2, m_2)$, you have $\log(\frac{m_1}{m_0}) \log(\frac{h_2}{h_0}) = \log(\frac{m_2}{m_0})\log(\frac{h_1}{h_0})$. There might be something simple this implies which I'm missing, but assuming a particular case of Schanuel's conjecture this implies that those logarithms are linearly dependent, which seems useful. – Command Master Dec 06 '23 at 04:37
  • Ah, okay. Thanks. – Zach Teitler Dec 06 '23 at 05:27
  • 1
    check Lucia's comment here https://mathoverflow.net/q/332532/4312 – Fedor Petrov Dec 06 '23 at 06:45
  • @FedorPetrov Thanks for the reference. I added an answer there that expands on Lucia's comment. Essentially the same argument works here. – Joshua Stucky Dec 18 '23 at 21:30

0 Answers0