<s>[INST] Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
   $$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to prove that is multiple of $5$?
 [/INST] Here is an answer to the actual question: why is $n^{19}-n^7$ a multiple of $5$ ?
Divide $n^{19}-n^7$ by $n^5-n$ as polynomials:
$$
n^{19} - n^7 = (n^{14} + n^{10} + n^6) (n^5 - n)
$$
By Fermat's theorem, $n^5-n$ is always a multiple of $5$ and the result follows.
Somewhat surprisingly, $n^5-n=n (n - 1) (n + 1) (n^2 + 1)
$ is always a multiple of $6$ and so the argument above proves that $n^{19}-n^7$ is always a multiple of $30$.
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