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I was looking at the resolution of the issue on this link (Find all positive integers $n$ such that $n$ divides $a^{25} - a$ for all positive integers $a$), but I have been trying for hours to understand why testing for 2 and 3 already guarantees the result for all "a" integers.

I had to make another post because I can't comment on posts yet. I can delete the post once my question is over if you want.

Thanks in advance

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Trying $a=2$ and $a=3$ only gives you an "upper bound" of which $n$ are possible (and fortunately a very "sharp" upper bound). Namely, if $n\mid a^{25}-a$ for all integers $a$, then necessarily we have $n\mid 2^{25}-2$ and $n\mid 3^{25}-3$, so it must be the case that $n\mid\gcd(2^{25}-2,3^{25}-3)$, i.e., $n\mid 2\cdot3\cdot 5\cdot 7\cdot 13$. In order to show that all these $n$ in fact do have the desired property, you may use Fermat.

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  • $\begingroup$ but, for example, wouldn't I be able to test for another number and not have 13 (for example)? $\endgroup$ – Helen Aug 29 '19 at 5:40
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    $\begingroup$ "but, for example, wouldn't I be able to test for another number and not have 13 (for example)? " Not sure what you mean. If you test for $n=11$ we don't have $11|a^{25}-a$ if $a = 2$. If you test for $n =17$ we do have $17|a^{25}-a$ for $a = 2$ but we don't have $17|a^{25}-a$ if $a = 3$. $\endgroup$ – fleablood Aug 29 '19 at 5:47
  • $\begingroup$ can't there be a number "a" such that, for example, we don't have $13 | a ^ {25} - a $? $\endgroup$ – Helen Aug 29 '19 at 6:14
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    $\begingroup$ @Helen: By the little theorem of Fermat, $a^{13}=a\pmod{13}$ and in consequence $a^{25}=a^{13}⋅a^{12}=a⋅a^{12}=a^{13}=a\pmod{13}$, so that indeed that can not happen. $\endgroup$ – Lutz Lehmann Aug 29 '19 at 9:34

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