Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3

Question

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.

Answer 1

Hint $\ $ Show $\ 3,5,8\mid f(n)\! =\! (n-2)(n-1)n(n+1)(n+2)$

implies that $\ 3,5,8\mid\, f(n+1)\,\ =\,\ (n-1)n(n+1)(n+2)(n+3)$

Answer 2

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}

Answer 3

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.

Answer 4

${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$

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