Is this a valid proof for $n^{5} - n$ is divisible by 30?

Asked May 27 '23 at 01:22

Active May 27 '23 at 02:04

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I know this question was asked in this thread How to prove n^5−n is divisible by 30 without reduction but my reputation is too low to comment. Here's how I approached it:

By Fermat's Little Theorem: $n^{p-1} \equiv\ 1 (\textrm{mod}\ p)$ where p is prime. 30 can be factored as (5)(6). $n^5 - n$ can be rewritten as $n(n^4 - 1)$. Therefore, my proof for 30 divides $n^5 - n$ is as follows:

$n(n^{4} -1)\equiv\ n(1 -1) \equiv\ 0 (\textrm{mod}\ 5(6))$.

Is this valid? All the comments in the other thread to something that seems overly complicated.

edited May 27 '23 at 01:26

asked May 27 '23 at 01:22

colinBerk

3

Welcome to Math SE. Note that Fermat's little theorem states that $n^{p-1}\equiv 1\pmod{p}$ for primes $p$ occurs only where $n$ doesn't contain any factors of $p$. Nonetheless, $n^{p}\equiv n\pmod{p}$ is true for all integers $n$ and primes $p$. – John Omielan May 27 '23 at 01:29
Format only applies to primes, $5\times 6$ isn't prime. And to apply it you need $n$ prime to $p$ – lulu May 27 '23 at 01:30
It might help that n^4 - 1 is evenly divisible by n-1. – Dan Asimov May 27 '23 at 02:02

Is this a valid proof for $n^{5} - n$ is divisible by 30?

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