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How do we show $$n^{97}\equiv n\text{ mod }4501770$$ for all integer $n$? First of all, I thought I could use Fermat's little theorem or Euler's theorem, but I'm not sure if they are applicable here.

0xbadf00d
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    You have to work one prime at a time. For example, $n^{96} \equiv 1$ (mod $5$) if $n$ is not divisible by $5,$ so $n^{97}-n$ is always divisible by $5.$ – Geoff Robinson May 31 '14 at 11:25

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As $\displaystyle4501770 =2\cdot3\cdot5\cdot7\cdot13\cdot17\cdot97$

Using Fermat's Little Theorem prime $p$ always divides $(n^p-n)=n(n^{p-1}-1)$

For example, $\displaystyle p=17, n^{97}-n=n(n^{96}-1)=n\{(n^{16})^6-1^6\}$ which is a multiple of $n(n^{16}-1)$ which is divisible by $17$ for all integer $n$

Similarly for the rest of the factors (which are also prime)

Now as $n^{97}-n$ is divisible by $2,3,5,7,13,17,97;$

$n^{97}-n$ will be divisible by their LCM which is their product here(right? why?)

Regarding how $96$ is identified, what is the LCM$(2-1,3-1,5-1,7-1,13-1,17-1,97-1)$

See also : Carmichael Function and calculate $\lambda(4501770)$

lab bhattacharjee
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    So to summarise: it all comes down to the fact that $m=4501770$ is a squarefree number with the property that for any prime $p$, if $p\mid m$ then $p-1\mid m-1$. – David May 31 '14 at 11:47
  • @David, In fact, it does not has to be. For example $8$ always divides $\displaystyle n^3(n^2-1);9|n^2(n^6-1)$ – lab bhattacharjee May 31 '14 at 11:51
  • @labbhattacharjee The elements in $\left{2,3,5,7,13,17,97\right}$ are pairwise coprime, so their LCM is their product. – 0xbadf00d May 31 '14 at 13:14
  • @oxbadfood, Nice to hear that. See also : http://math.stackexchange.com/questions/652126/proof-that-a5-b-b5-a-is-divisible-by-30-for-any-integers-a-and-b – lab bhattacharjee May 31 '14 at 13:17
  • @labbhattacharjee so in your comment you are asserting, correctly, that the converse of my comment is not true. – David May 31 '14 at 14:39
  • @David, What do you mean by converse? – lab bhattacharjee May 31 '14 at 14:41
  • If the conditions in my comment are true then the OP's result is true. You have correctly stated that if the conditions in my comment are false then the OP's result need not be false. – David May 31 '14 at 14:49
  • @labbhattacharjee One last question: We've got $$4501770;|;(n^{97}-n)\Leftrightarrow n^{97}\equiv n\text{ mod }4501770$$ So, there's nothing more to show. But you've asked me to consider $$\text{lcm}\left{2-1,3-1,5-1,7-1,13-1,17-1,97-1\right}=96=\lambda (4501770)$$ I can see that there's a useful relation between these things, but how does it help initially (before we've shown the divisibility properties) in this question? – 0xbadf00d May 31 '14 at 14:56
  • @oxbadfood, Any problem is best understood by its origin. Here is does not have to be $97$. it can be $$n^{96a+1}-n$$ right? – lab bhattacharjee May 31 '14 at 14:58