<s>[INST] Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.

If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.

 [/INST] Let $n = 2k$ for some integer $k$. Then,
$$\begin{align}2k((2k^2) + 20) &= 2k(4k^2 + 20)\\
&=8k(k^2 + 5)\end{align}$$
But 
$$\begin{align}k(k^2 + 5) &\equiv k(k^2-1)\mod 2\\
&=(k+1)(k)(k-1) \mod 2\end{align}$$
Since $2$ must divide at least one of these three consecutive integers, we have $2 \mid k(k^2 + 5) \implies 16 \mid 8k(k^2 + 5)$.
Similarly, 
$$\begin{align}k(k^2 + 5) &\equiv k(k^2-1)\mod 3\\
&=(k+1)(k)(k-1) \mod 3\end{align}$$
Using the same argument, $3$ must divide at least one of these three consecutive integers. Hence $3 \mid k(k^2 + 5) \implies 3 \mid 8k(k^2 + 5)$.
Since both $3$ and $16$ divides $n(n^2 + 20)$, we get that $48$ must also divide $n(n^2 + 20)$.
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