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From Dirichlet Pigeonhole Principle if $p$ is a prime and if $a,b\in\mathbb Z$ are in $(0,p/2)$ then there is a $t\in(0,p)\cap\mathbb Z$ such that $\|(x,y)\|_\infty<\lceil\sqrt p\rceil$ holds where $t(a,b)\equiv(x,y)\bmod p$.

  1. Is there no distinct coprime $a,b$ in $(0,\lceil\sqrt p\rceil)$ such that there is $t\in(0,p)\cap\mathbb Z$ with $\|(x,y)\|_\infty<\|(a,b)\|_\infty$?

From Dirichlet Pigeonhole Principle if $p$ is a prime and if $a,b,c\in\mathbb Z$ are in $(0,p/2)$ then there is a $t\in(0,p)\cap\mathbb Z$ such that $\|(x,y,z)\|_\infty<\lceil p^{2/3}\rceil$ holds where $t(a,b,c)\equiv(x,y,z)\bmod p$.

  1. Is there no distinct pairwise coprime $a,b,c$ in $(0,\lceil p^{1/3}\rceil)$ such that there is $t\in(0,p)\cap\mathbb Z$ with $\|(x,y,z)\|_\infty<\|(a,b,c)\|_\infty$? Is it possible to increase size of $a,b,c$ to a larger value than $\lceil p^{1/3}\rceil$?
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  • $\begingroup$ Do you mean for $t$ in your question to also lie in $(0,p)$? If so, then $(a,b)=(1,1)$ trivially implies the answer is "no" $\endgroup$ – Wojowu Dec 24 '18 at 13:41
  • $\begingroup$ Let $(a,b)=(2,2)$ then. For $t=(p+1)/2$, $(x,y)=(1,1)$. $\endgroup$ – Wojowu Dec 24 '18 at 13:45
  • $\begingroup$ $(a,b)=(2,4),t=(p+1)/2,(x,y)=(1,2)$ $\endgroup$ – Wojowu Dec 24 '18 at 13:47
  • $\begingroup$ @Wojowu coprime $a,b$. $\endgroup$ – Freeman. Dec 24 '18 at 13:52
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No. Observe that $a/b=x/y$, from which it follows that $ay=xb$. Since $a,b,x,y$ are all in $(0,\sqrt{p})$, it follows that the equality $ay=xb$ holds not merely mod $p$, but in $\mathbb{Z}$. Since $a,b$ are coprime, it follows that $a\mid x$ and $b\mid y$. Hence, $\lvert x\rvert\geq \lvert a\rvert$ and $\lvert y\rvert\geq \lvert b\rvert$.

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  • $\begingroup$ If $p$ is not a prime does something like this hold at least when $p$ is square free odd and has only $O(1)$ prime factors? $\endgroup$ – Freeman. Dec 25 '18 at 8:54

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