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Let $p$ be a prime. $|a| = min \{\bar{a},p-\bar{a}\}$ be the Lee norm. Let

$\chi(a) = 0$ if $\bar{a}=0(p)$

$\chi(a) = +1$ if $2\bar{a}\le p$

$\chi(a) = -1$ if $2\bar{a}> p$

Define $\frac{\partial a}{\partial p} := \frac{a-\chi(a)|a|}{p}$, $a':= \sum_{p|a, p \text{prime}}\frac{\partial a}{\partial p}= a \sum_{p|a, p \text{prime}} \frac{1}{p}$

So far I have been able to show for example (using (5)):

$p,q>2$ primes then: $\frac{\partial}{\partial q}\frac{\partial a}{\partial p} = \frac{\partial}{\partial p} \frac{\partial a}{\partial q} $

Here are a few questions concerning this arithmetical "derivatives":

(1) $(ab)' \le a'b+b'a$ equality holds if and only if $gcd(a,b)=1$

(2) $a \le b$ then $\frac{\partial a}{\partial p} \le \frac{\partial b}{\partial p}$

(3) $\frac{\partial a}{\partial p}

(4) $\chi(a) \equiv \frac{a}{|a|} \text{ mod } p$ for $a \neq 0 \text{ mod } p$

(5) Let $p>2$ then: $\frac{\partial a}{\partial p} = \left \lfloor \frac{a+\frac{p-1}{2}}{p} \right \rfloor$

(6) If $a,b \ge \frac{p+1}{2}, p>2$ then: $\frac{\partial ab}{\partial p}

(7) Let $p>2$. Then we have: $\frac{\partial a+b}{\partial p} = \frac{\partial a}{\partial p}+\frac{\partial b}{\partial p}$ if and only if $|a+b|=|a|+|b|+2k$, $0 \le k \le \frac{p-1}{2}$

(8) The generating function of the sequence $\frac{\partial n}{\partial p}$ where $n\ge 1$ is $\frac{x^{\frac{p+1}{2}}}{(x-1)^2\phi_p(x)}$ where $\phi_p(x)$ is the $p$-cyclotomic polynomial.

If someone has an idea how to prove one of the questions, that would be very nice.

Thanks for your help!

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    Could you please cite the source of this question. – 2017-02-17
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    Dear @BillDubuque: This is a question out of curiosity. The source is, that I did a lot of computer experiments to derive these questions. – 2017-02-17

1 Answers 1

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Here is a strong hint for $(4)$:

Note that $2\overline{a} \leq p \Rightarrow |a| = \overline{a}$ and that $2\overline{a} > p \Rightarrow |a| = p-\overline{a}$. With the definition of $\chi(a)$ and the observation that $\frac{a}{|a|} \equiv \frac{\overline{a}} {|\overline{a}|} \mod p$, this can be used to prove $(4)$.