Exercise 2.1.23 ($n^{13} - n $ is divisible by $2,3,5,7$ and $13$)

Question

Prove that $n^{13} - n $ is divisible by $2,3,5,7,$ and $13$ for any integer $n$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Note that
\begin{align*}
n^{13} - n &= n(n^{12}-1)\
&= n (n^6 - 1) (n^6 + 1)\
&= n(n^3 - 1) (n^3 + 1) (n^6 + 1)\
&=n(n-1)(n^2 + n + 1) (n^3 + 1)(n^6 + 1).
\end{align*}
The complete factorization is 
$$n^{13} - n = n(n-1)(n^2 + n + 1) (n+1)(n^2 - n + 1) (n^2 + 1)(n^4 - n^2 + 1).$$
By Fermat's theorem, $2 \mid n^2 -n = n(n-1) \mid n^{13}-n$, thus
$$2 \mid n^{13} -n.$$
Similarly, $3 \mid n^3 - n = n(n-1)(n+1) \mid  n^{13} - n$, thus
$$3 \mid n^{13} -n.$$
Moreover $5 \mid n^5 - n = n(n-1)(n+1)(n^2 + 1) \mid n^{13} - n $, thus
$$5 \mid n^{13} -n.$$
Likewise, $7 \mid n^7 -n = n(n^3-1)(n^3 + 1) \mid n^{13}-n$, thus
$$7 \mid n^{13} -n.$$
Finally, by Fermat's theorem
$$13 \mid n^{13} -n.$$
$n^{13} - n $ is divisible by $2\cdot 3 \cdot 5 \cdot 7 \cdot 13= 2730$, for every integer $n$.
\end{proof}

Exercise 2.1.23 ($n^{13} - n $ is divisible by $2,3,5,7$ and $13$)

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Exercise 2.1.23 ($n^{13} - n $ is divisible by $2,3,5,7$ and $13$)

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