Determine the greatest common divisor of the elements of the set $S = \{ n^{13} - n \mid n \in \mathbb{Z} \}.$
I put in a few values of $n$ and I know that $10$ is a common factor, but I'm not sure if that's the greatest common factor.
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2See here. It is $2730=2\cdot 3\cdot 5\cdot 7\cdot 13$. – Dietrich Burde Jun 04 '20 at 13:48
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1For the general case, see https://math.stackexchange.com/questions/164524/largest-modulus-for-fermat-type-polynomial – lhf Jun 04 '20 at 13:55
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3Does this answer your question? How to show that $2730\mid n^{13}-n;;\forall n\in\mathbb{N}$ – Mike Earnest Jun 06 '20 at 02:13
2 Answers
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By Fermat, $n^{13} \equiv n \mod p$ if $p-1$ divides $12$. Since the divisors of $12$ are $1,2,3,4, 6, 12$, the corresponding primes are $2, 3, 5, 7, 13$. And indeed the gcd of $2^{13}-2$ and $3^{13}-3$ is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$.
Robert Israel
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By Fermat's Little Theorem, we know that $13\mid n^{13}-n, 7\mid n^7-n, 5\mid n^5-n,3\mid n^3-n$ and $2\mid n^2-n$. I claim the product $2\cdot3\cdot5\cdot7\cdot13=2730$ is the largest common factor.
Note that $2^{13}-1=8190=3\cdot2730$ and $(3^{13}-3)-(2^{13}-1)=2730\cdot581$ where $\gcd(3,581)=1$ so $2730$ is the answer.
hdighfan
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