<s>[INST] Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.

I was thinking of using induction, but wasn't really sure how to do it.
 [/INST] If one wishes to use induction:
The statement holds for $n=0$, so let us assume it's true for $n=k$: $$k^2(k^2+1)(k^2-1)=(k-1)k^2(k+1)(k^2+1)=5m $$ and consider it for $n=k+1$. We have $$(k+1)^2((k+1)^2+1)((k+1)^2-1)=\\ (k+1)^2(k^2+2k+2)(k^2+2k)=\\ k(k+1)^2(k+2)(k^2+2k+2).$$ Suppose this isn't a multiple of $5$. Then none of its factor is, so our inductive hypothesis yields that for some integer $a,b$ either $k-1=5a$ or $k^2+1=5b$, or both: 


*

*if both, then $$(k-1)+(k^2+1)=k(k+1)=5(a+b),$$ contradiction;

*if only the former, we have $$k+4=5(a+1),$$ whence $$ k^2+2k+2=25(a+1)^2-6k-14=25(a+1)^2-2(3k+7)=\\25(a+1)^2-2(3(k-1)+10)=5\left(5(a+1)^2-2\left(\frac{3(k-1)}{5}+2\right)\right),$$ contradiction;

*if only the latter, we also have for some integer $c$ $$k+3=5c,$$ whence $$k^2+2k+2=k^2+2k+7-5=(k^2+1)+2(k+3)-5=5(b+2c)-5=5(b+2c-1),$$ contradiction.

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