<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] Say $$n \equiv 0,\pm1,\pm2 \pmod 5$$ or, 
$$n^2 \equiv 0,1,4 \pmod 5$$ or,
$$n^6 \equiv 0,1,64 \pmod 5$$ or, 
$$n^6 \equiv 0,1,-1 \pmod 5$$
Therefore, $5$ divides at least one of $n^6,n^6-1$ or $n^6+1$, that is, $5$ divides $n^6(n^6-1)(n^6+1)$ .
And 
$n^{19}-n^7=n\cdot n^6(n^{12}-1)=n\cdot n^6(n^6-1)(n^6+1)$
Hence given expression is divisible by $5$ and thus by $30$.
Hope this helps.
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