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This is a question on the remainders of integer division from my student.

Notations. Let $p$ be a positive odd prime integer. We write $r_{i,j}$ for the remainder of $i \times j \div p$. Now for an integer $u \in \{1, 2, \dots, p-2\}$, we define \begin{align*} K_1 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u<p,\ r_{i,p-u-1} <u+1\}, \\ K_2 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u<p,\ r_{i,p-u-1} >u+1\}, \\ K_3 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u>p,\ r_{i,p-u-1} <u+1\}, \\ K_4 =&\{ i : 1 \le i \le p-2,\ r_{i,u}+u>p,\ r_{i,p-u-1} >u+1\}. \end{align*} And for another $u' \in \{1, 2, \dots, p-2\}$, we can define $K_\nu'$ for $\nu=1,2,3,4$.

Question. If $K_1 =K_1'$ and $K_4=K_4'$, then either $K_2=K_2'$ and $K_3=K_3'$, or $K_2=K_3'$ and $K_3=K_2'$.

What I have done.

1. With the help of computer, I checked the correctness of this question for $p<100$.

2. Still with the help of computer, I found that, at least for $p<100$, if $K_1 =K_1'$ and $K_4=K_4'$, then either $u \equiv u' \pmod{p}$ or $u+u' \equiv p-1 \pmod{p}$. The first case corresponds to the result $K_2=K_2'$ and $K_3=K_3'$, and the second one corresponds to the result $K_2=K_3'$ and $K_3=K_2'$.

3. I observed the definitions of the $K_\nu$'s, and found some equivalent conditions. Firstly, we have \begin{align*} r_{i,u}+u<p &\quad\Leftrightarrow\quad r_{i+1,u} =r_{i,u} +u,\\ r_{i,u}+u>p &\quad\Leftrightarrow\quad r_{i+1,u} =r_{i,u} +u-p, \end{align*} and similarly, \begin{align*} r_{i,p-u-1} <u+1 &\quad\Leftrightarrow\quad r_{i+1,p-u-1} =r_{i,p-u-1} +(p-u-1),\\ r_{i,p-u-1} >u+1 &\quad\Leftrightarrow\quad r_{i+1,p-u-1} =r_{i,p-u-1} +(p-u-1) -p. \end{align*}

4. Using the discussion in (3), we obtain \begin{align*} i \in K_1 &\;\Leftrightarrow\; r_{i,u}+r_{i,p-u-1}<p \text{ but } r_{i+1,u}+r_{i+1,p-u-1}>p, \\ i \in K_4 &\;\Leftrightarrow\; r_{i,u}+r_{i,p-u-1}>p \text{ but } r_{i+1,u}+r_{i+1,p-u-1}<p, \\ i \in K_2 \cup K_3 &\;\Leftrightarrow\; \text{both $r_{i,u}+r_{i,p-u-1}$ and $r_{i+1,u}+r_{i+1,p-u-1}$ are $<p$ or $>p$}. \end{align*}

5. Now by the result in (4), we can see that $K_1 =K_1'$ and $K_4=K_4'$ if and only if for every $i \in \{1, 2, \dots, p-1\}$, the signatures of $r_{i,u}+r_{i,p-u-1}-p$ and $r_{i,u'}+r_{i,p-u'-1}-p$ are the same (i.e., both positive or both negative).

However, I really do not know how to solve this question, so if you have any idea, please help me. Thank you very much for your attention.


Any ideas or hints are welcome. Please.

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1 Answer 1

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To continue with your argument: you have the property that $r_{i,u}+r_{i,p−u−1}−p$ and $r_{i,u′}+r_{i,p−u′−1}−p$ have the same sign for all $i$. This is equivalent to saying that $r_{i,u} - r_{i,u+1}$ and $r_{i,u'} - r_{i,u'+1}$ have the same sign for all $i$. This difference is always $p-i$ or $-i$, so this means that $$ r_{i,u} - r_{i,u+1} = r_{i,u'} - r_{i,u'+1} $$ for all $i=1,2,\ldots,p-2$.

One can also argue that this means that for all $i$ the sum of fractional parts $$ \{\frac{iu}p\} + \{\frac{i(u'+1)}p\} +\{\frac{i(-u')}p\} =1+\{\frac{i(u+1)}p\}. $$ This becomes a statement that some toric cyclic singularity in dimension three is terminal. These have been classified. I believe that corollary 1.3 of "TERMINAL QUOTIENT SINGULARITIES IN DIMENSIONS THREE AND FOUR" DAVID R. MORRISON AND GLENN STEVENS http://www.ams.org/journals/proc/1984-090-01/S0002-9939-1984-0722406-4/S0002-9939-1984-0722406-4.pdf should give you the desired statement.

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  • $\begingroup$ Thank you very much for your generous help. $\endgroup$
    – KWSK
    May 27, 2015 at 0:13

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