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 $n(n^2+20)$.

 A: 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.
A: 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
$$
m(m^2+5)\equiv m(m^2+2)\equiv m^3+2m\equiv m+2m=3m\equiv 0\pmod{3}.
$$
A: $48=2^4\cdot 3$, 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
A: ${} \bmod 16$, 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 $48=16\cdot 3=lcm (16,3)$.
A: $48 = 16 \cdot 3$
It's easy to see (as amWhy pointed out) that $16$ divided $n(n^2 + 20)$
(Note that if $n = 2k$, then $n(n^2+20) = 8k(k^2+5)$, but one between $k$ and $k^2 + 5$ must be even, so we have divisibility by $16$)
Also, 
if $n \equiv 0 \pmod 3$, then also $n(n^2 + 20)$ is divible by $3$ and we're done.
If $n \equiv 1, 2, \pmod 3$, then $n^2 + 20 \equiv 21 \equiv 0 \pmod 3$, and we're done.
A: 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 $n(n^2 + 20)$.
A: This is a different way of looking at some of the other answers. Let $n=2m$ then $$n(n^2+20)=8m(m^2+5)=8\left(m(m^2-1)+6m\right)=48\binom {m+1}3+48m$$ gives an explicit expression as an integer multiple of $48$.
A: Here is another approach , we can use $-4 + 24 = 20$
$$ n(n^2 + 20) = n(n^2 - 4 + 24)$$
$$n(n + 2)(n-2) + 24n$$
n is even = 2k for integer k
$$2k(2k + 2)(2k - 2) + 48k = 2k2(k+1)2(k-1) + 48k$$
$$ 8k(k+1)(k-1) + 48k$$
Note $(k-1)k(k+1) = 6m$ is the product of 3 consecutive integers and therefore divisible by $6 = 3!$ 
$$ 8 \cdot 6m + 48k = 48(m + k)$$
for integers m , k. This shows $n(n^2 + 20) $ is a multiple of 48 therefore divisible by 48.
A: Why don't you try principal of mathematical induction here.
Just assume that n=2m for a natural numbers m. Now you get a new expression which is like $${n(n^2+20) = 8m(m^2+5)}$$
If we just try induction here, it is clear that P(1) is true. Suppose that P(K) is true, and you will get that $${8K(K^2+5) = 48T}$$ for some positive integer T.
And then we just need to prove P(K+1) is true, which I shall let you to try.
