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For fixed nonnegative integers $a$ and $b$ such that $a>b$, let $$g(a,b):=\underset{n\in\mathbb{Z}}{\gcd}\,\left(n^a-n^b\right)\,.$$ Here, $0^0$ is defined to be $1$. (Technically, we can also set $g(a,b):=0$ if $a=b$, and $g(a,b):=g(b,a)$ if $a<b$.) For a nonzero integer $m$ and a prime integer $p$, let $v_p(m)$ denote the maximum power $k\in\mathbb{N}\cup\{0\}$ such that $p^k$ divides $m$. The set $D(a,b)$ is given by $$D(a,b):=\big\{p\in\mathbb{N}\,\big|\,p\text{ is an odd prime and }p-1\mid a-b\big\}\,.$$ (a) Show that $$g(a,b)=\left\{\begin{array}{ll}1\,,&\text{if }b=0\text{ and }a\text{ is odd}\,, \\ 2\,,&\text{if }b>0\text{ and }a-b\text{ is odd}\,, \\ \displaystyle 2^{\min\big\{v_2(a-b)+2,b\big\}}\,\prod_{p\in D(a,b)}\,p^{\min\big\{v_p(a-b)+1,b\big\}}\,,&\text{if }a-b\text{ is even}\,.\end{array}\right.$$ (b) What is the minimum value $\ell(a,b)$ of $l\in\mathbb{N}\cup \{0\}$ such that there exists a subset $S\subseteq\mathbb{Z}$ of size $l$ for which $$g(a,b)=\underset{n\in S}{\gcd}\,\left(n^a-n^b\right)\,?$$ (With the convention $\gcd(\emptyset)=0$, we can say that $\ell(a,b)=0$ for $a=b$.)

For examples, $$g(a,0)=1$$ for all integers $a>0$, $$g(13,1)=2\cdot 3\cdot 5\cdot 7\cdot 13=2730\,,$$ and $$g(8,2)=2^{2}\cdot 3^2\cdot 7=252\,.$$ This post is inspired by Prove that $2730$ divides $n^{13} - n$ for all integers $n$.. Part (a) is known. Part (b) is still open, and is related to The GCD of a Univariate Integer-Valued Polynomial over a Set. We have the following bound: $$\ell(a,b)\leq a+1\,.$$ However, I believe that this inequality holds: $$\ell(a,b)\leq 2\,.$$

Batominovski
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    Part $(a)$ has something to do with the carmichael function, but I could not work out the details yet. – Peter Jul 05 '16 at 11:02
  • and did you get anywhere with this too? – it's a hire car baby Nov 13 '18 at 10:42
  • @RobertFrost Unfortunately, no, to both problems you asked about. – Batominovski Nov 13 '18 at 20:16
  • Do you know how to prove that if $p$ is an odd prime such that $p\not\in D(a,b)$, then $p\not\mid g(a,b)$ ? (This might be obvious, but I cannot prove that rigorously. I tried to solve this problem and almost got $G=\Gamma$, but had a similar doubt there.) – mathlove Jul 18 '20 at 11:41
  • @mathlove First, if $p\notin D(a,b)$, then there exists $t\in\mathbb{Z}$ such that $p\nmid t$ and $t^{a-b}\not\equiv 1\pmod{p}$. This is because the greatest common divisor of the polynomials $x^{a-b}-1\in\mathbb{F}_p[x]$ and $x^{p-1}-1\in\mathbb{F}_p[x]$ is the polynomial $$x^{\gcd(a-b,p-1)}-1$$ of degree smaller than $p-1$. Therefore, the roots of $x^{\gcd(a-b,p-1)}-1$ cannot be all the elements of $\mathbb{F}_p\setminus{0}$. Then, $p\nmid t^b\big(t^{a-b}-1\big)=t^a-t^b$. – Batominovski Jul 18 '20 at 11:48
  • I got it, thanks. (I've found an error in my argument, so it is difficult for me to find $G$. If I'm not mistaken, all I've got so far is the followings which I think you should already notice : (1) $v_2(G)$. (2) For $n$ such that $n\equiv 0\pmod p$, $v_p(f(n))\ge k+a$. (3) For $n$ such that $n\not\equiv 0\pmod p$, $v_p(f(n))\ge\sum\limits_{\substack{{1\leq i\leq k}\{p-1|b_i-1}}},\big(1+v_p\left(b_i-1\right)\big)$. (4) $v_3(G)$. What I can't prove is the existence of $n$ for which the equality in the inequality (3) is attained. If we can find such an $n$ for every $p$, then $G=\Gamma$.) – mathlove Jul 19 '20 at 11:26

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