<s>[INST] 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.
 [/INST] ${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$

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