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Take $p$ to be a prime and let $a_1,\dots,a_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{ma_1}p,\dots,\frac{ma_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D$.

Then we know there is an integer $t\in\{1,\dots,p\}$ such that the vectors $$\{\frac{ta_1}p,\dots,\frac{ta_n}p\}$$ are in any box of size $c\mathcal D$ for some constant $c>0$.

This means for any box of size $((c\mathcal D)^{1/n}p)^n$ in $[-p/2,p/2]^n$ there is an integer $t$ such that $$t(a_1,\dots,a_n)\bmod p$$ lies in that box. For example there is a $t$ such that $t(a_1,\dots,a_n)\bmod p$ is in $[2(cD)^{1/n}p,3(cD)^{1/n}p]\times [0,(cD)^{1/n}p]^{n-1}$.

  1. Is it possible to reduce the box size from $((c\mathcal D)^{1/n}p)^n$ to something smaller?

Take $p$ to be a prime and let $a_1,\dots,a_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{ma_1}p,\dots,\frac{ma_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D_a$ and let $b_1,\dots,b_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{mb_1}p,\dots,\frac{mb_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D_b$.

  1. What is the smallest size such that given any such box we have $t_a,t_b\in\{1,\dots,p-1\}$ such that $$(t_a(a_1,\dots,a_n)+t_b(b_1,\dots,b_n))\bmod p$$ lies in that box?
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