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$2730=2\cdot3\cdot5\cdot7\cdot13$

$n^{13}-n=n(n-1)(n+1)(n^2+1)(n^8+n^4+1)$

Divisibility by $2$ and $3$ follows from the product of two and three successive terms.

Divisibility by $5$ follows from Fermat's little theorem: $$n^{5-1}\equiv1(\mod 5)\Rightarrow 5|(n^{13}-n)=(n^4-1)(n^8+n^4+1)$$

How to prove divisibility by $7$ and $13$?

user300045
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  • This question is asked see http://math.stackexchange.com/questions/596074/how-to-show-that-2730-mid-n13-n-forall-n-in-mathbbn?rq=1 – Adi Dani Nov 20 '16 at 10:22
  • You know Fermat's little theorem but don't apply it to $13$ and $7$? Why? – Arthur Nov 20 '16 at 10:25

2 Answers2

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  • You have also by Fermat's little theorem :

$$n^{13}\equiv n\pmod{13}$$

because $13$ is prime.

  • Be careful, you have $n(n^4-1)(n^8+n^4+1)=n^{13}-n$, I think you made a typo.

  • Dr Sonnhard Graubner treat the case $7$ perfectly !

E. Joseph
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note that $$n \equiv 0,1,2,3,4,5,6 \mod 7$$ then $$n^{13}\equiv 0,1,2,3,4,5,6 \mod 7$$

Dr. Sonnhard Graubner
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