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For my homework, I need to prove that prove that $n^{7}−n$ is divisible by 42 for any integer n.

Here is the part of my proof where I use Fermat's:

$$ \\\textbf{Claim 1: }7\vert n^{7} - n\\ n^{7} - n = n(n^{6} - 1)\\ n^{6}\equiv_71\text{, by Fermat's Little Theorem}\\ n(1-1) \textbf{mod}7 = 0\\ \implies 7\vert n^{7} - n\\ $$

I am assuming that if you have $a\cdot b\cdot c (\textbf{mod} d)$ then you can replace a (WLOG) with (a $\textbf{mod}$ d).

My logic is that you are finding the remainder when you divide $a\cdot b\cdot c $ by d, and you can split that up into $\frac{a}{d}\cdot \frac{b}{d}\cdot \frac{c}{d} $

Is this correct?

banana
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    I would say that "$n^p\equiv n\pmod{p}$ for every integer $n$" is the full version of Fermat's Little Theorem. Your proof seems to ignore the case $7\mid n$. (But fortunately, there is a factor $n$ there.) – Batominovski Oct 13 '18 at 19:21
  • @lulu why is it not the case that because 7 is a divisor of 42 that 42 is a divisor of n^7-n? I understand it examples, but what's the actual explanation? – banana Oct 13 '18 at 19:24
  • @Batominovski Hm. Does the case of $7\vert n$ come about when we subtract n from $n^{7}$? I was taught in class that fermat's theorem was that $n^{p-1} \equiv_p 1$ for prime p – banana Oct 13 '18 at 19:26
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    In your class, there has to be the assumption that $p\nmid n$. Otherwise, $n^{p-1}\equiv 1\pmod{p}$ is clearly false for all integers $n$. – Batominovski Oct 13 '18 at 19:27
  • @Batominovski Sorry, just to clarify. What makes this clear? I am new to proof based math... I really appreciate your willingness to answer, though. – banana Oct 13 '18 at 19:32
  • It is very clear because $0^{p-1}\not\equiv 1\pmod{p}$ for any prime $p$. This is because $p\mid 0$. – Batominovski Oct 13 '18 at 19:34
  • @Batominovski Ah thank you so much!! People like you really help students like me who otherwise struggle with these concepts :) I appreciate it! – banana Oct 13 '18 at 19:35
  • Here is a related, more general, thread: https://math.stackexchange.com/questions/1849603/. It shows that $$\gcd_{n\in\mathbb{Z}}\left(n^7-n\right)=42,.$$ (Even more generally, see this deleted thread: https://math.stackexchange.com/questions/2384486. I hope that it would be reopened soon.) – Batominovski Oct 14 '18 at 08:36

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I suggest working with
$n^7 - n = n(n^3 - 1)(n^3 + 1) = n(n^3 - 1)(n^2 - n + 1)(n + 1).$

William Elliot
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