Consider the subring $\text{Int}(\mathbb{Z})=\{f\in\mathbb{Q}[x]:f(m)\in\mathbb{Z}\text{ for all }m\in\mathbb{N}\}$. Show $2$ is irreducible but not prime in $\text{Int}(\mathbb{Z})$.
I was able to show $2$ is irreducible in $\text{Int}(\mathbb{Z})$ by writing $2=fg$ for some non-unit constant polynomials $f,g\in\text{Int}(\mathbb{Z})$, and plugging in $1$ to both sides. This gives $2=f(1)g(1)=fg$ where $f,g\in\mathbb{Z}$. Since $2$ is irreducible in $\mathbb{Z}$ and the units of $\mathbb{Z}$ coincide with the units of $\text{Int}(\mathbb{Z})$, we obtain a contradiction. However, I am not sure about showing $2$ is not prime. I need to find two elements of $\text{Int}(\mathbb{Z})$ whose product is divisible by $2$, but neither factor is divisible by $2$. I am sure that these factors will have to be some kind of higher degree polynomials, but I am scratching my head here. One thing that might be helpful, is the fact that $$ \binom{x}{n}=\frac{x(x-1)\cdots(x-n+1)}{n!}\in\text{Int}(\mathbb{Z}), $$ but I am not sure how this applies here, if at all (it was mentioned in the problem statement). Any help is appreciated.
It's simple enough to see that for all integers $x$, $x(x-1)$ is even, but $x$ and $x-1$ are not divisible by $2$ because they take on odd values. However, $x(x-1)$ is divisible by $2$, since $$x(x-1)=2\binom {x}2$$ Intuitively, you can factor $2$ out of a polynomial that always takes even values on integers, because it will result in a polynomial that takes integer values on integers. Its constituents in the product need not always take on even values though, as long as at least one is even for each integer.