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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 $\mathrm{mcm}(q_1,...,q_{n-1})$.\mathrm{lcm}(q_1,...,q_{n-1})$. Any idea for the general case? 4 added 21 characters in body; added 8 characters in body 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, and I have the impression that that might not be the case : if some of the$t_j$'s are irrational. t_j=\frac{p_j}{q_j}$, then it suffices to choose $q\equiv 1$ mod $\mathrm{mcm}(q_1,...,q_{n-1})$.

Any idea for the general case?

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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, and I have the impression that that might not be the case if some of the $t_j$'s are irrationalsirrational. Any idea?

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