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

  • 1
    $\begingroup$ 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


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.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.