I have tried to show that : $2730 |$ $n^{13}-n$ using fermat little theorem but i can't succeed or at a least to write $2730$ as $n^p-n$ .
My question here : How do I show that $2730$ divides $n^{13}-n$ for $n$ is integer ?
Thank you for any help
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$2730=2\times 3\times 5\times 7\times 13$ and apply Little Fermat to each prime factor – marwalix Apr 09 '16 at 09:40
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Then of course Chinese Remainder theorem... – marwalix Apr 09 '16 at 09:47
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See also http://math.stackexchange.com/questions/164524/largest-modulus-for-fermat-type-polynomial. – lhf Apr 09 '16 at 09:48
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Hint: Use Fermat's Theorem, working separately modulo $2$, $3$, $5$, $7$, and $13$. (Note that $2730=2\cdot 3\cdot 5\cdot 7\cdot 13$.)
André Nicolas
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First you calculate the prime factorisation of $2730$. You will find that it splits in 5 prime factors. Then use the Chinese Remainder theorem and show that $n^{13}\equiv n \pmod{p}$ for the prime factors.
Quentchen
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You can use Fermat's little theorem. It does not matter if $n$ is greater than $p$. – Quentchen Apr 09 '16 at 19:38