Proof of Divisibility of $n(n^2+20)$ by 48.

Since $n$ is even, we can write $n = 2k$ for some integer $k$.

Hint:

$$n(n^2 + 20) = 2k((2k)^2 + 20)= 2k(4k^2 + 20) = 8k(k^2 + 5)$$ Hence, $8$ is a factor.

Note further that one of $k$ or $k^2 + 5$ must be even, and hence divisible by $2$. Why?

So now we know that $8\cdot 2 = 16$ is a factor.

All that remains to be shown is that $3$ is also a factor.

The question is equivalent to proving that, for any integer $m$, $6$ divides $m(m^2+5)$, because for $n=2m$ the expression is $8m(m^2+5)$.

Divisibility of $m(m^2+5)$ by $2$ is obvious, because $m^2\equiv m\pmod{2}$, so $$ m(m^2+5)\equiv m(m+1)\equiv m^2+m\equiv 2m\equiv 0\pmod{2}. $$ Divisibility of $m(m^2+5)$ by $3$ follows similarly, because $m^3\equiv m\pmod{3}$, so $$ m(m^2+5)\equiv m(m^2+2)\equiv m^3+2m\equiv m+2m=3m\equiv 0\pmod{3}. $$

$48=2^4\cdot 3$, 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