Using the fact that $n^3 + 5n$ is divisible by $6$, infer that $n^4 - 2n^3 + 11n^2 + 14n$ is always divisible by $24$

Question

The problem is as stated in the title. I've tried factorization, but it doesn't seem to yield anything. Any help is greatly appreciated. Also please don't use modular arithmetic, thank you.

Answer 1

Hint: Note that \begin{align} n(n-1)(n-2)(n-3)&=n^4-6n^3+11n^2-6n=\color{red}{[n^4-2n^3+11n^2+14n]}-4\color{blue}{[n^3+5n]}. \end{align}

Answer 2

We have $$P(n)=n^4-2n^3+11n^2+14n=n(n+1)(n^2-3n+14)$$

►Reducing modulo $3$ $$P(n)\equiv n(n+1)(n^2-1)=(n-1)(n-1)n(n+1)\equiv 0\pmod 3$$ because has a factor which is product of three consecutive integers.

►Reducing modulo $8$ $$P(n)\equiv n(n+1)(n^2+5n+6)=n(n+1)(n+2)(n+3)\equiv 0\pmod 8$$ because is a product of four consecutive integers.

Thus $P(n)$ is divisible by $24$ for all $n$.

NOTE.- By the way, $n^3+5n$ is always divisible by $6$ (reduce modulo 3).

Answer 3

Alternatively $\,\displaystyle f(n) = (n\!+\!1)n(n\!-\!1)(n\!-\!2) + 12(n\!+\!1)n = \color{#c00}{24}{n+1\choose 4}+ \color{#c00}{24}{n+1\choose 2}$

Answer 4

You don't need to use that $n^3 + 5n$ is divisible by $6$: $$ n^4 - 2n^3 + 11n^2 + 14n = 24\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}\right) $$