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I saw a proof of why 42 | $n^7-n$ here

But I was a little confused. By Fermat's Little Theorem for any prime $p$ and any $a \in {(1,2,..., p−1)}$, we have $a^{p−1}$ ≡ $1$ $($mod p$)$ .

So, don't we need to split into two cases, where $n \geq 7$ and $n < 7$? Because we don't know what $n$ is, and the def of Fermat only applies if $n$ is within the set $(1,2,3,4,5,6,7)$?

Bill Dubuque
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Manny
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    If 7|(n⁷-n) and 6|(n⁷-n) then 42|(n⁷-n) as 6,7 are.relatively prime – user600016 Sep 26 '20 at 01:55
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    start over and prove divisibility by $2$ and $3$ and $7$ – Will Jagy Sep 26 '20 at 01:57
  • Do you even know what FLT is ? Because after knowing FLT it will be easy for you to deduce that $7$ divides $n^7 - n$ . What you have written is that $a^{p-1} \equiv 1$ $($mod p$)$ , how can you rewrite it? – Anonymous Sep 26 '20 at 03:52

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You have a confused view of what Fermat's little theorem states. It says that $p|a^p-a$ regardless of $a$.

$a^p-a$ can be factored into $a(a^{p-1}-1)$ so if it happens to be the case that $p\nmid a$ then $p|a^{p-1}-1$ as long as $a$ isn't a multiple of $p$.

Your list of numbers for $a$, where it is the case that $p|a^{p-1}-1$ wouldn't stop at $p-1$. It would skip $p$ and would include all numbers from $p+1$ to $2p-1$, then skip $2p$ and include all numbers from $2p+1$ to $3p-1$, then skip $3p$ and include all numbers from $3p+1$ to $4p-1$, etc. Basically the set includes all natural numbers that are not divisible by $p$.

The division of the problem between $n\lt 7$ and $n\ge 7$ makes no sense. The division of the problem would be between $7|n$ and $7\nmid n$.

quantus14
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