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Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$.

$n^7 \equiv n \bmod 2\implies n^6 \equiv 1 \bmod 2$

Using Fermat's little theorem, it is easy to see that:

$n^1 \equiv 1\bmod 2\implies n^a \equiv 1^a\bmod 2, \forall a \in \mathbb{N}$

Algebraic approach: $(n^3-1)(n^3+1)\equiv 0\bmod 2$. So, if either of the two factors is divisible by $2$, then fine. So, now have:

  1. $n^3 \equiv 1 \pmod 2$
  2. $n^3 \equiv -1 \pmod 2$

So, how to pursue? Cannot get anything by stating $n^3 = 1 + 2k, k\in \mathbb {Z}$, or $n^3 = -1 + 2l, l\in \mathbb{Z}$.

Nor, will gain anything by assuming a form of $n$.

jiten
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    If $n \equiv 0 \bmod 2$, then $n^7 \equiv 0 \equiv n \bmod 2$; and if $n \equiv 1 \bmod 2$, then $n^7 \equiv 1 \equiv n \bmod 2$. Fermat’s Little Theorem gives us that $a^{p-1} \equiv 1 \bmod p$ when $a$ is not a multiple of $p$. – azif00 Jun 29 '22 at 03:49
  • @azif Request Answer, as that might have helped more by details, or even insightful comments to that. – jiten Jun 29 '22 at 03:59
  • @azif Is your approach of considering both cases is similar to answer here (https://math.stackexchange.com/a/2433004/424260)? – jiten Jun 29 '22 at 04:11
  • Easy: use parity: $,\forall n!:, 2 \mid f(n) !\iff! \forall n!:, 2\mid f(n\bmod 2)!\iff 2\mid f(0)\ \ &\ \ 2\mid f(1),,$ for any poly $f(x)$ with integer coef's. For more general methods see the linked dupes – Bill Dubuque Jun 30 '22 at 07:10
  • @BillDubuque it does not seem at all to be a duplicate of the question which you've marked it as being a duplicate of. – Suzu Hirose Jun 30 '22 at 07:16
  • @SuzuHirose If you read the linked posts you will learn that the answers include most of the common methods used to handle problems of this type, including those in the answers below. – Bill Dubuque Jun 30 '22 at 07:20

2 Answers2

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If $n$ is odd then $n^7$ must be odd, so $n^7-n$ must be even. If $n$ is even then $n^7$ is even, so $n^7-n$ must be even.

Suzu Hirose
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 30 '22 at 07:10
  • @BillDubuque - duly noted, I'll check the "Related" bar before answering next time. – Suzu Hirose Jun 30 '22 at 07:13
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    That usually won't suffice. Instead either search for dupes or at least delay answering Qs that are clearly very likely to be dupes. – Bill Dubuque Jun 30 '22 at 07:17
  • @BillDubuque Here is a similar question I found via the related bar: https://math.stackexchange.com/questions/2432993/prove-that-forall-n-in-bbb-n3-mid-n7-n?noredirect=1&lq=1 but you haven't yet marked it as a duplicate, so I guess that it may be useful after all. – Suzu Hirose Jun 30 '22 at 07:21
  • There are hundreds of possible dupe targets for this FAQ. I don't see anything new there. – Bill Dubuque Jun 30 '22 at 07:22
  • I found a duplicate which you hadn't found using the bar, so it might be useful to look there. – Suzu Hirose Jun 30 '22 at 07:23
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$n^7-n = n(n^6-1) = n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1)$. Observe the product consists of $2$ consecutive integers $n,n-1$ hence divisible by $2$.

Wang YeFei
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