Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.
 A: Using repeated differences and Newton's interpolation formula  we get
$$
n^5-5n^3+4n = 120 \binom{n}{3} + 240 \binom{n}{4} + 120 \binom{n}{5} 
$$
Although this identity suffices for answering the question, it also implies the simpler identity below:
$$
n ^5-5n^3+4n = 120 \binom{n+2}{5}
$$
which gives a crystal clear answer to the question.
If you must use induction, then:
\begin{align}
f(n+1)-f(n)
&=5 n^4+10 n^3-5 n^2-10 n\\
&= 5 (n+2) (n+1) n  (n-1)\\
&= 5 (4!) \binom{n+2}{4}\\
&= 120 \binom{n+2}{4}
\end{align}
A: Let it be true for k
$$k^3(k^2-1) - 4k(k^2-1) = k(k^2-4)(k^2-1) = (k+2)(k+1)k(k-1)(k-2)$$ is divisible by 5,4,3,2 and hence 120
Now replace k by k+1
Then $$(k+3)(k+2)(k+1)k(k-1)$$ is still divisible by 5,4,3,2 and hence 120
It is true for any n>=3.
A: Hint $\ $ Show $\ 3,5,8\mid f(n)\! =\! (n-2)(n-1)n(n+1)(n+2)$ 
implies that $\ 3,5,8\mid\, f(n+1)\,\ =\,\ (n-1)n(n+1)(n+2)(n+3)$
A: ${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$

*

*${n^5-5n^3+4n = (n-2)(n-1)(n)(n+1)(n+2)=  5!\binom{n+2}{5}}$
