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I have been asked to find $n$ satisfying

$n^{17} \equiv n\;\;$(mod $4080$)

I'm truly unsure how to even begin to attempt this. Is there a theorem or property of congruences that I should be identifying as useful for this problem?

rainingagain
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  • Try the Chinese remainder theorem – lhf May 05 '22 at 12:20
  • Isn't that used to solve systems of congruences, as opposed to single problems? How could it be applied here? – rainingagain May 05 '22 at 12:22
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    $4080=2^4\cdot 3\cdot 5\cdot 17$. You can solve it modulo each of those prime power factors (keeping Fermat's little theorem and Euler's theorem in mind), and then combine the solutions with the chinese remainder theorem. – Jaap Scherphuis May 05 '22 at 12:34
  • Sorry, I'm not sure what you mean by saying solve it module each of those factors. Do you mean solving 4 separate instances of $n^{17} \equiv n;$(mod $2^{4})$, $n^{17} \equiv n;$(mod $3)$, and so on? How is this done using Fermat's little theorem? (Sorry but I'm not used to solving congruence problems involving n on both sides of the relation). – rainingagain May 05 '22 at 12:41
  • Yes, that is what I mean. What does Fermat/Euler's theorem modulo 5 say? What does that tell you about $n^{17} \pmod{5}$? Which values for $n$ therefore satisfy $n^{17}\equiv n \pmod 5$ ? – Jaap Scherphuis May 05 '22 at 12:48
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    See also Carmichael's lambda function, noting that $\lambda(4080)=16$. – lhf May 05 '22 at 13:50

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By Fermat's Little Theorem, for $p \nmid n$, $n^{p-1}\equiv 1\pmod p$ and so $n^{k(p-1)+1}\equiv n \pmod p$ holds for all $n$ (as it holds trivilally for multiples of $p$).

Note that $4080=2^4\cdot3\cdot5\cdot17$. By our initial observation, it follows that the congruence $n^{17}\equiv n$ holds modulo $p$ for $p=3,5 $ or $17$. Note that, modulo $16$, if $n$ is even the congruence holds iff $16|n$, and if not, then $n^8\equiv1\pmod {16}$ (By Fermat-Euler or just noticing that fourth powers of odd numbers are $1$ modulo $16$), whence $n^{17}\equiv n\pmod {16}$.

Finally, by chinese remainder theorem, it follows that $n^{17}\equiv n \pmod {16\cdot3\cdot5\cdot17}$ holds iff $n$ is odd or a multple of $16$.

Anubhab
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