<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] ${} \bmod 16$, we have $n(n^2+20) \equiv n(n^2+4)=2k(4k^2+4) = 8k(k^2+1)$. If $k$ is even then clearly $8k(k^2+1)\equiv 0$. If $k$ is odd, then $k^2+1$ is even and again $8k(k^2+1)\equiv 0$.
${} \bmod 3$, we have $n(n^2+20) \equiv n(n^2-1)=(n-1)n(n+1)\equiv 0$, since given three consecutive numbers, exactly one of them is a multiple of $3$.
Therefore, $16$ and $3$ divide $n(n^2+20)$ and so does $48=16\cdot 3=lcm (16,3)$.
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