1
$\begingroup$

Let $p$ a natural number greater than $5$, $H$ a subgroup of $(\mathbb Z/p\mathbb Z)^*$.

Is it true that : if $a\in H$ with $a>2$ and $c=-\frac{1}{p} \mod a$

then $(a−1)|\sum \limits_{h\in H} p(h\times c \mod a)$ ?

Source : les dattes à Dattier

  • 0
    What do you mean with $a>2$? Remember that $a$ is a residue class, not a number. – 2017-01-24
  • 0
    $a\neq 0 \text{ and } a\neq 1$ – 2017-01-24
  • 0
    I identifie $\mathbb Z/ p\mathbb Z$ to the number set {0,1, 2,...,p-1} – 2017-01-24
  • 0
    $a \mod p$ is the remainder of the euclidean division of a by p. – 2017-01-24
  • 0
    sorry I forget $a\neq 2$, but unimportant here – 2017-01-24
  • 0
    What do you mean by $-h/p\mod{a}$? Is that a division and modulo? Why would $h/p$ be always integer? Or why would $p$ be invertible in $\mod{a}$ if we are talking about division modulo? The right side is very unclear to me. – 2017-02-10
  • 0
    pgcd(p,a)=1 because $a\in(\mathbb Z/p\mathbb Z)^*$ so p is invertible mod a, and $(1/p \mod a)\in \{1,..,a-1\}$ – 2017-02-10
  • 0
    No, the sum is in $(\mathbb N,+)$ not in $\mathbb Z/p \mathbb Z$ – 2017-02-12
  • 0
    $(h\mod a)\in\mathbb N\cap [0,a−1]$ – 2017-02-12
  • 0
    For subgroups $H$ that contain $p-1$, this is because $H$ can be partitioned into disjoint sets of the form $\{h,p-h\}$, and $(p-h)c\bmod a = a - 1 - (hc\bmod a)$, so the sum of the two mods is $a-1$. But in general, $H$ need not contain $p-1$. – 2017-04-01
  • 0
    But in general, H need not contain p−1 : yes – 2017-04-01

1 Answers 1

4

Since $a \in H$, the mapping $h \mapsto (ah \text{ mod } p)$ is a permutation on $H$. We write this map as

$$ (ah \text{ mod } p) = ah - \lfloor ah/p\rfloor p. $$

Now let $S$ denote the sum in question. Using the fact that $(x \text{ mod } a) = (y \text{ mod } a) $ whenever $x \equiv y \pmod{a}$, we can write

\begin{align*} S = \sum_{h \in H} p (hc \text{ mod } a) &= \sum_{h \in H} p ((ah \text{ mod } p)c \text{ mod } a) \\ &= \sum_{h \in H} p (( ah - \lfloor ah/p\rfloor p)c \text{ mod } a) \\ &= \sum_{h \in H} p (\lfloor ah/p\rfloor \text{ mod } a). \end{align*}

But since $0 \leq \lfloor ah/p \rfloor < a $, it follows that $(\lfloor ah/p\rfloor \text{ mod } a) = \lfloor ah/p\rfloor$. Plugging this back,

$$S = \sum_{h \in H} p \lfloor ah/p\rfloor = \sum_{h \in H} ah - \sum_{h \in H} (ah \text{ mod } p). $$

Utilizing the permutation $h \mapsto (ah \text{ mod } p)$ again, this simplifies to

$$S = (a-1)\sum_{h\in H} h $$

and therefore the claim follows.

  • 0
    Bravo, It's impressive for me – 2017-04-02
  • 0
    Can I propose you a last enigme ? – 2017-04-02
  • 0
    Good for being a "weak point"! :) – 2017-04-02
  • 0
    I don't understand @mickep – 2017-04-02
  • 1
    In another (linked) question Sanghul Lee mentioned that his/her weak point was algebra. – 2017-04-02
  • 1
    My strong points seems weaker than its weak points... lol – 2017-04-02
  • 0
    Can you give the solution (which can be give in one line) to this question : http://math.stackexchange.com/questions/2213224/a-general-result-find-a-sequence-of-c2-convex-function-uniformly-convergent @Sangchul Lee, you would help me. – 2017-04-02
  • 0
    Now, It is a problem of degree 8 difficulty @Sangchul Lee – 2017-04-02
  • 1
    @mickep, Haha, thank you. To be honest, this idea was a pure luck that is far above my algebra skills. And you may call me 'he' as both the gravatar and my name suggest. (Indeed, it is a typical masculine name in my country.) – 2017-04-03
  • 0
    @Sangchul Lee : Every time I find a solution, it's as if I win at loto, I see the winning draw formed in my head, without I do anything else but want to find the solution. Thanks My God. – 2017-04-03