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 $n(n^2+20)$.


$48=2^4\cdot 3$, so we will work modulos $16$ and $3$.
We have $n(n^2+20)=n^3+20n$. Now in modulo 3, one notices that $n^3\equiv n\pmod 3$ for any n, so that $n^3+20n\equiv n+20n\equiv 21n\equiv 0\pmod 3$. So we're done with divisibility by 3.
We now need to show divisibility by 16. We have $n(n^2+20)\equiv n(n^2+4)\pmod {16}$. Since $n$ is even, let $n=2k$. We must show $2k(4k^2+4)\equiv 0\pmod {16}$, or $8k(k^2+1)\equiv 0\pmod {16}$. This is equivalent to $k(k^2+1)\equiv 0\pmod 2$, which is obviously true since $\gcd(k,k^2+1)=1$.
Now since $\gcd(3,16)=1$, and $3\mid n(n^3+20)$ and $16\mid n(n^2+20)$, it follows $48=3\cdot 16\mid n(n^2+20)$ and we're done. $\blacksquare$
Arkan
