Find all the numbers $a$ such that the number $an(n+2)(n+4)$ is an integer for all $n \in \mathbb{N}$
It's trivial to see that if $a$ is irrational, we get no solution.
Thus $a \in \mathbb{Q} \Rightarrow a = \dfrac{p}{q} ~ (*)$ where $p, q \in \mathbb{Z}$ and $\gcd(p,q)=1$. Then
$$\begin{split} an(n+2)(n+4)=k &\stackrel{(*)}{\Rightarrow} \dfrac{p}{q}n(n+2)(n+4)=k \\ & \Rightarrow pn^3+6pn^2+8pn-kq=0 \end{split}$$
By the integer root theorem, for $n$ to be an integer it should be $kq = zn ~ (1)$. Hence
$$\begin{split} \Rightarrow pn^3+6pn^2+8pn-kq=0 &\stackrel{(1)}{\Rightarrow} pn^2+6pn+8p-z=0 \\ &\Rightarrow n = \dfrac{-6p \pm 2\sqrt{p^2+zp}}{2p} \\ &\stackrel{n>0}{\Longrightarrow} n = -3 + \dfrac{\sqrt{p^2+zp}}{p} \in \mathbb{Z}\end{split}$$
But again for $n$ to be an integer it should be $p^2+zp=p^2m^2 \Rightarrow p = \dfrac{z}{m^2-1} ~(2)$
By (1) and (2) we get that $$kq=p(m^2-1)n \Rightarrow \dfrac{p}{q} = \dfrac{k}{(m^2-1)n} \stackrel{n=m-3}{\Longrightarrow} \boxed{a = \dfrac{k}{(m-3)(m-1)(m+1)}}$$
We see that if $m$ is a small number (like $2$ and $3$) the equation is satisfied, but for $m$ being a larger number (like $10$)then the initial number is not an integer. Please tell me if I have done anything wrong or if I should add anything to my solution.
Considering $n=1$, $n=2$ we see that necessarily $15a\in\Bbb Z$ and $48a\in\Bbb Z$. But then already $3a=48a-3\cdot 15a\in\Bbb Z$. On the other hand, if $3a\in \Bbb Z$ then $an(n+2)(n+4)\in\Bbb Z$ for all $n$ because the numbers $n,n+2,n+4$ are pairwise incongruent modulo $3$ and hence one of them is a multiple of $3$. We conclude that $a$ has the desired property if and only if $$a\in\frac13\Bbb Z.$$