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 $5$?

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 $\color{red}{30}$ 
