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 $5$?
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.
Notice, one should re-write & factorize as follows $$n^{19}-n^7=n^7(n^{12}-1)$$ $$=n\cdot \underbrace{n^6\color{blue}{(n^{6}-1)}\color{red}{ (n^{6}+1)}}_{\text{divisible by 5}}$$ $$=n^7\color{blue}{(n^{3}-1)(n^3+1)}\color{red}{(n^{2}+1)(n^4-n^2+1)}$$
$$=n^7\color{blue}{(n-1)(n^2+n+1) (n+1)(n^2-n+1)}\color{red}{(n^{2}+1)(n^4-n^2+1)}$$ $$=n^4\underbrace{(n-1)n(n+1)}_{\text{divisible by 3!}}\ \underbrace{n^2(n^2+1)}_{\text{divisible by 2! }}(n^8+n^4+1)$$ thus it clear that $\color{blue}{n^{19}-n^7}$ is divisible by $5\times 3!\times 2!=60$ i.e. it is divisible by $\color{red}{30}$
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 $30$.
