Consider $0=t_0\leq t_1\leq...\leq t_n=1$, $f_0,...,f_{n-1}\in\mathbb{Z}$ and $F:[0,1]\to\mathbb{R}$ be such that

1) $F\equiv f_i$ on the interval $(t_i,t_{t+1})$, for all $i=0,...,n-1$,

2) $\displaystyle \int_0^1 F(t) dt=\sum_{i=0}^{n-1}(t_{i+1}-t_i)f_i=0$.

Does there exist an arbitrarily large prime number $p$ and a positive integer $k=k(p)$ such that $q:=p^k$ satisfies

$\displaystyle \sum_{i=1}^{q-1} F\left(\frac{i}{q}\right)=0$ ?

I know that the answer is YES when all the $t_j$'s are rational number: if $t_j=\frac{p_j}{q_j}$, then it suffices to choose $q\equiv 1$ mod $\mathrm{lcm}(q_1,...,q_{n-1})$.

Any idea for the general case?

  • $\begingroup$ Are the two $n$'s in your problem the same? $\endgroup$ Commented Dec 2, 2011 at 16:30
  • $\begingroup$ Sorry, that was a typo. $\endgroup$ Commented Dec 2, 2011 at 16:32
  • 2
    $\begingroup$ Is it known whether there are infinitely many integers $m$ such that $\sum_{i=1}^{m-1} F(i/m) = 0$? $\endgroup$ Commented Dec 2, 2011 at 19:55
  • 1
    $\begingroup$ @Greg It's not known to me. I would be happy with that already. $\endgroup$ Commented Dec 2, 2011 at 22:25
  • 1
    $\begingroup$ @GH: I think you might be able to improve your proof to the case where q is prime, using results about equidistribution of primes. The Theorem of Green-Tao quoted in these lecture notes looks like it should do the job, but I haven't checked the original reference yet, austms.org.au/tiki-download_file.php?fileId=162 (Horocycle Flow at Prime Times) $\endgroup$ Commented Dec 3, 2011 at 2:24

1 Answer 1


Thanks to the comments of George Lowther and Greg Martin (for which I am most grateful), I can now show that the answer is YES for infinitely many primes $q$.

Theorem 1. Let $t_1\dots,t_{n-1}$ be any finite set of real numbers.Then for any $\epsilon>0$ and any integer $r>0$ there are infinitely many primes $q\equiv 1\pmod{r}$ such that $$\|(q-1)t_i\|<\epsilon,\qquad i=1,\dots,n-1.\tag{1}$$ Here $\|x\|$ stands for the distance of $x$ to the nearest integer.

Proof. Without loss of generality, the numbers $1,t_1,\dots,t_{n-1}$ are linearly independent over $\mathbb{Q}$. Indeed, we can express each of them as a $\mathbb{Z}$-linear combination from a suitable basis $\frac{1}{s},t_1^*,\dots,t_{m-1}^*$ of their $\mathbb{Q}$-linear span, where $s>0$ is an integer. Then the statement for $t_1,\dots,t_{n-1}$ follows from the statement for $t_1^*,\dots,t_{m-1}^*$ (with $\mathrm{lcm}(r,s)$ in place of $r$). When the elements of $1,t_1,\dots,t_{n-1}$ are linearly independent over $\mathbb{Q}$, the statement follows from the stronger result that the vectors $(qt_1,\dots,qt_{n-1})$ are dense modulo $1$, as $q$ runs through the primes such that $q\equiv 1\pmod{r}$. This result is essentially due to Vinogradov, with technical improvements by Vaughan and Harman. See Theorem 4 in Harman: Diophantine approximation with primes, J. London Math. Soc. (2) 39 (1989), 405–413. Well, Harman does not have the condition $q\equiv 1\pmod{r}$, but it seems straightforward to incorporate it.

Theorem 2. Let $t_i$, $f_i$, $F$ be as in the original question but without the assumption $f_i\in\mathbb{Z}$. Then for any $\epsilon>0$ there is a prime $q$ such that $$\left|\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)\right|< \epsilon.$$ In particular, if $F$ is integer valued and $\epsilon=1$, then the left hand side is zero.

Proof. Assume that $\epsilon>0$ is sufficiently small, namely $$\epsilon<\|t_i\|,\qquad i=1,\dots,n-1.\tag{2}$$ By Theorem 1, there is a prime $q$ such that (1) holds. Observe that $$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=0}^{n-2}f_i\bigl([qt_{i+1}]-[qt_i]\bigr)+f_{n-1}\bigl(q-1-[qt_{n-1}]\bigr),$$ where $[x]$ stands for the integral part of $x$. Here we used that no $\frac{j}{q}$ coincides with any $t_i$, as follows from (1) and (2). We subtract $$0=(q-1)\int_0^1 F(t)dt=\sum_{i=0}^{n-1}f_i\bigl((q-1)t_{i+1}-(q-1)t_i\bigr),$$ then with the notation $$\tilde t_i:=[qt_i]-(q-1)t_i,\qquad i=0,\dots,n-1,$$ we get $$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=0}^{n-2}f_i\bigl(\tilde t_{i+1}-\tilde t_i\bigr)-f_{n-1}\tilde t_{n-1}=\sum_{i=1}^{n-1}(f_{i-1}-f_i)\tilde t_i.$$ By (1) we can write $(q-1)t_i=n_i+e_i$ with $n_i\in\mathbb{Z}$ and $|e_i|<\epsilon$. Then $$qt_i = n_i+t_i+e_i,\qquad i=1,\dots,n-1,$$ whence by (2), that is by $\epsilon<\min(t_1,1-t_{n-1})$, we have $[qt_i]=n_i$, so that $\tilde t_i=-e_i$. It follows that $$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=1}^{n-1}(f_i-f_{i-1})e_i,$$ whence $$\left|\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)\right|<\epsilon\sum_{i=1}^{n-1}|f_i-f_{i-1}|.$$ The right hand side can be made arbitrary small, so we are done.

  • 2
    $\begingroup$ Nice job GH. I think it follows from general exponential sum techniques that there are infinitely many primes $p=q-1$ that satisfy (2). (One probably gets an asymptotic formula even, although the constant will depend upon the linear relations among the $t_i$ in a perhaps complicated way.) $\endgroup$ Commented Dec 3, 2011 at 5:31
  • $\begingroup$ @George: You are right, George Lowther made a similar comment below. I will update my answer in a moment. Thank you. $\endgroup$
    – GH from MO
    Commented Dec 3, 2011 at 6:33
  • $\begingroup$ Thanks a lot GH!!! In Hamiltonian dynamics, this theorem should implies the following: if the average Maslov index of an orbit $\gamma$ is zero, then for infinitely many prime numbers $p$ the $p$-iteration of the orbit gamma has the same Maslov index as $\gamma$. Maybe one may get something useful out of this... $\endgroup$ Commented Dec 4, 2011 at 4:41
  • $\begingroup$ @Marco: Sounds interesting, although I am not familiar with this theory. I am glad I could help! $\endgroup$
    – GH from MO
    Commented Dec 4, 2011 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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