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Let $p$ be a prime integer and $n\geq 3$ be a natural number such that $p^n-1$ has a primitive prime divisor. Also let $n=\lambda_1+\cdots+\lambda_m$ such that $n>\lambda_i\geq 1$ for all $i$. I want to show that $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ does not divide $p^n-1$.

This is my suggested argument for proving the statement when $p>2$:
Both of $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ and $p^n-1$ are polynomials in $p$ with the same degree. If $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)\mid p^n-1$, we then have $p^n-1=c(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$, where $c\in \mathbb{N}$. Since $p>2$, $p^{\lambda_i}-1>1$ for all $i$. Therefore, the coefficient of $p^{\lambda_1+...+\lambda_m}=p^n$ in the expansion of the product $\prod_{i=1}^{m}{(p^{\lambda_i}-1)}$ is 1. Also the coefficient of $p^n$ in $p^n-1$ is $1$, and so we must have $c=1$ . Therefore $p^n-1=(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$. Now if we consider $q$ to be primitive prime divisor of $p^n-1$, then $q$ does not divide $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$, which is a contradiction. So if $p>2$, it is impossible for $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1)$ to divide $p^n-1$.

But as it was discussed in comments, it seems that my argument is not true.
Question 1: Does my argument work? If not, is there any possible way to prove the statement for $p>2$?
The remaining case is when $p=2$. According to the example mentioned in comments $(2-1)(2^2-1)(2^3-1) \mid 2^6-1$.
Question 2: Is there any example of a pair $(n, p)\neq(6,2)$ and a partition $\lambda=(1\leq\lambda_1\leq...\leq\lambda_m<n)$ of $n$ such that $p^n-1$ has a primitive prime divisor and $(p^{\lambda_1}-1)\cdots(p^{\lambda_m}-1) \mid p^n-1$? (Update: There are plenty of such pairs when $p=2$, see comments of Nate below).
I would be grateful for any help.

Math_Student
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    Have you tried when all $\lambda_i$'s are $1$? – Batominovski Aug 15 '16 at 14:26
  • What is special about the case when all $\lambda_i=1$? I still can't see whether $(p-1)^n$ divides $p^n-1$ or not... – Math_Student Aug 15 '16 at 14:34
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    It is not a special case. If you want to know whether the answer to your question is 'yes' or 'no,' it may be a useful first step to try out trivial cases. – Batominovski Aug 15 '16 at 14:39
  • @Batominovski That's right! If $(p-1)^n \mid p^n-1$, then we have $p^n-1=c \Sigma_{i=0}^{n}{(-1)^i \binom{n}{k}p^{n-k}}$, where $c<p$. Since $c<p$, multiplying $ \Sigma_{i=0}^{n}{(-1)^i \binom{n}{k}p^{n-k}}$ by $c$ does not change any powers of $p$. Hence we have some powers of $p$ in the right-hand which are not in the left hand. So $(p-1)^n$ does not divide $p^n-1$. Am I right? – Math_Student Aug 15 '16 at 15:07
  • @Batominovski Can we extend this argument to the general case where $p^n-1$ has a primitive prime divisor? – Math_Student Aug 15 '16 at 15:35

3 Answers3

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This is not true in general. Take $p=5$ and $n=3$. Then $\frac{(5^3-1)}{(5-1)^3}=\frac{31}{16}$.

Pax
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It doesn't happen in general, but... $(2^1-1)(2^2-1)(2^3-1) \vert (2^6-1)$

Nate
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  • What do you mean by "in general"? Does the divisibility happen if we choose $n$ to be a perfect number and $\lambda_i$'s be proper divisors of $n$ (as your example)? – Math_Student Aug 15 '16 at 16:04
  • I just meant that it is not true that $(p^\lambda_1-1) \dots (p^\lambda_m-1) | (p^n-1)$ for all choices of $p$, $n$, and $\lambda_i$ as in the problem, but that here is an example where it does hold. – Nate Aug 15 '16 at 19:02
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    I'll also note that for $p=2$ you can find lots of examples by including a bunch of terms where $\lambda_i = 1$, since multiplying by $1$ won't change divisibility. – Nate Aug 15 '16 at 19:05
  • Thank you. That's right. What about the case $p>2$? Is my argument for $p>2$ correct? – Math_Student Aug 15 '16 at 19:17
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    No, I don't think your solution is correct. You seem to be conflating the notions of a polynomial dividing another polynomial, and a number dividing another number. Given two polynomials $f(x)$ and $g(x)$ with integer coefficients, it is perfectly possible for $f(n)$ to divide $g(n)$ at some values of $n$ even if $f(x)$ does not divide $g(x)$ as polynomials. – Nate Aug 15 '16 at 19:32
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    Also: $(3^1-1)(3^1-1) | (3^2-1)$ – Nate Aug 15 '16 at 19:34
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    That's right. By the way, all of your examples are contained in the pairs $(n, p)$ such that $p^n-1$ does NOT have any primitive prime divisor. So is possible to prove the non-divisibility for the cases that primitive prime divisors exist? – Math_Student Aug 15 '16 at 19:44
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Hint: you can also prove that $gcd(p^a-1, p^b-1) = p^{gcd(a,b)}-1$ like in this question.

Community
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Quang Hoang
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  • Would you please explain more about the relation of your hint to the question? As far as I understood, the fact in your hint implies that $\lambda_i \mid n$ for all $i$. – Math_Student Aug 15 '16 at 16:00