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Let $p, q$ be two distinct prime number. I'm trying to provide a non-trivial upper bound for the sum $$S(p, q) = \sum_{1 \leq x < p} \sum_{1 \leq y < q} \frac{1}{\|x / p\| \, \|y / q\| \, \|x/p + y/q\|},$$ where $\|t\|$ denotes the distance of $t \in \mathbb{R}$ from the nearest integer.

Precisely, I'm interested in having $S(p, q) = o((pq)^2)$ as $p, q$ go to infinity in some way.

I know that $\min(p, q) \to +\infty$ doesn't suffices, since $S(p, q) \geq (pq)^2 / (q - p)$ (considering $x = 1$ and $y = q - 1$), and we can take a sequence of primes $p_k < q_k$ such that $p_k \to +\infty$ and $q_k - p_k$ is bounded.

Maybe $S(p, q) = o((pq)^2)$ as $p \to +\infty$ and $q/p \to +\infty$ ?

The motivation comes from the fact that $$\sum_{\substack{1 \leq z < pq \\ (pq, z) = 1}} \frac1{|p^{-1}z \bmod q|\,|q^{-1}z \bmod p| \, z} = \frac{S(p, q)}{(pq)^2} ,$$ where $|p^{-1}z \bmod q| = \min\{|r| : r \in \mathbb{Z}, pr \equiv z \pmod q\}$, and similarly for $|q^{-1}z \bmod p|$. Therefore, $S(p, q) = o((pq)^2)$ means that, on average, $|p^{-1}z \bmod q|$, $|q^{-1}z \bmod p|$, and $z$ cannot be all small.

Thanks for any help

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    $\begingroup$ I'm fairly certain that one can obtain the bound $$S(p,q) \ll \frac{(pq)^2}{p+q}(\log p)(\log q).$$ If I have the time, I'll write up an answer to show this. Note that this is essentially sharp, since the term with $x=y=1$ is $$\frac{(pq)^2}{p+q}.$$ $\endgroup$ Dec 26, 2021 at 0:05

2 Answers 2

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If you just want $o()$, the story is rather simple. Let $a_{xy}$ be the remainder of $qx+py\mod pq$ where all remainders modulo $P$ are assumed to be between $-P/2$ and $P/2$. Note that all $a_{xy}$ are distinct, so if we have any set $Z$ of pairs $(x,y)$, then $\sum_{(x,y)\in Z}\frac 1{a_{xy}}\le 2(1+\log|Z|)$. What we want to show is just $$ \sum_{0<|x|<p/2, 0<|y|<q/2}\frac 1{|xya_{xy}|}=o(1)\,. $$ Now for $k=0,1,2\dots$ consider $Z_k=\{(x,y): 2^k\le |xy|<2^{k+1}\}$ and note that $|Z_k|\le C(k+1)2^{k}$. Thus the sum over $Z_k$ is at most $2^{-k}(1+\log|Z_k|)\le C(k+1)2^{-k}$ regardless of $p,q$. Thus the only danger is that the sum over $Z_k$ for some fixed $k$ does not tend to $0$, i.e., that there exists $C>0$ such that $ap+bq+c=0$ for some $a,b,c$ with $0<|a|+|b|+|c|<C$ along a subsequence of pairs $(p,q)$ you are considering (in which case the corresponding term alone gives a positive constant). If you eliminate this possibility in any way ($q/p\to+\infty$ is more than enough), then you are in good shape.

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This is not mean to be a full answer, but one which illustrates how one can prove an estimate like the one in my comment through a rather ``brute force'' approach.

To illustrate the idea of the computation, let's deal with the case when $p=2$ and $q$ is some odd prime number, so $$ S(2,q) = \sum_{1 \leq y < q} \frac{1}{\frac{1}{2} \left|\left| \frac{y}{q}\right|\right| \cdot \left|\left| \frac{1}{2} + \frac{y}{q}\right|\right|}. $$ Note that the expression inside the second $||\cdot||$ symbol in the denominator is always in $(\frac{1}{2},\frac{3}{2})$, and therefore $$ \left|\left| \frac{1}{2} + \frac{y}{q}\right|\right| = \begin{cases} \frac{1}{2} - \frac{y}{q} & \text{if $\frac{1}{2} + \frac{y}{q} < 1$}, \\ \frac{y}{q} - \frac{1}{2} & \text{if $\frac{1}{2} + \frac{y}{q} > 1$}. \end{cases} $$ This motivates us to decompose $S(2,q)$ as $$ S(2,q) = 2 \Big(\sum_{1\leq y \leq \frac{q-1}{2}} + \sum_{\frac{q+1}{2}\leq y < q} \Big)\frac{1}{ \left|\left| \frac{y}{q}\right|\right| \cdot \left|\left| \frac{1}{2} + \frac{y}{q}\right|\right|} = 2(T_1 + T_2), $$ say. In each sum, the $||\cdot||$ expressions can be evaluated explicitly, and we obtain $$ \begin{aligned} T_1 &= \sum_{1\leq y \leq \frac{q-1}{2}} \frac{1}{\frac{y}{q} \left(\frac{1}{2} - \frac{y}{q} \right)} = q^2 \sum_{1\leq y \leq \frac{q-1}{2}} \frac{1}{y\left(\frac{q}{2}-y \right)}, \\ T_2 &= \sum_{\frac{q+1}{2}\leq y < q} \frac{1}{\left(1-\frac{y}{q} \right) \left(\frac{y}{q} -\frac{1}{2} \right)} = q^2 \sum_{\frac{q+1}{2}\leq y < q} \frac{1}{(q-y)\left(y-\frac{q}{2}\right)}. \end{aligned} $$ Note that in each of the expressions in the denominators on the right, exactly one of the two factors is larger than $\frac{q}{4}$ (to see this, just split the sums depending as $y\leq \frac{q}{4}$ and $y\leq \frac{3q}{4}$, respectively). Omitting a few details of the calculations, we have $$ S(2,q) \ll q \sum_{1\leq y \leq \frac{q}{4}} \frac{1}{y} \ll q\log q. $$ There's some technicalities about changing variables in the second sum and what the exact limits of summation work out to be, but the above inequality is morally correct.

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  • $\begingroup$ For the general case $p > 2$, the computation gets really messy $\endgroup$
    – Seee
    Dec 26, 2021 at 14:28

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