Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$ I am a student and this question is part of my homework.
May you tell me if my proof is correct?
Thanks for your help! 
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$.
$(z^3-z)(z+2)=z(z^2-1)(z+2)=z(z-1)(z+1)(z+2)=(z-1)(z)(z+1)(z+2)$
$(z-1)(z)(z+1)(z+2)$ means the product of $4$ consecutive numbers.
Any set of $4$ consecutive numbers has $2$ even numbers, then $(z-1)(z)(z+1)(z+2)$ is divisible by $4$.
Any set of $4$ consecutive number has at least one number that is multiple of $3$, then $(z-1)(z)(z+1)(z+2)$ is divisible by 3.
Therefore $(z-1)(z)(z+1)(z+2)$ is divisible by $12$. Q.E.D.

That's correct. Alternatively it is divisible by $\,24\,$ by integrality of binomial coefficients
$$\,(z+2)(z+1)z(z-1)\, =\, 4!\ \dfrac{(z+2)(z+1)z(z-1)}{4!}\, =\, 24{ {z+2\choose 4}}\qquad\qquad$$
Similarly $\,n!\,$ divides the product of $\,n\,$ consecutive naturals.
