When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer $a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\left(b+x\right)}$$
And then I been trying to say:
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right) $$
$$a+m\left(b+x\right)| (a+m\left(b+x\right))(b+x)$$
So
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right)+ (a+m\left(b+x\right))(b+x)$$
$$a+m\left(b+x\right)| m\left(\left(b+x\right)^2+b^2\right)$$
But I don't see how it helps, so please help me.

For every even $x>0$ we have that $f(x)$ is not an integer for all $a,b,m$. To see this, chose an even $x$ and take $a=b=m=x$. Then 
$$
f(x)=\frac{-x^3}{-2x^2-x}=\frac{x^2}{2x+1}.
$$
Because $x^2$ is even, and $2x+1$ is odd, the fraction cannot be an integer. Similarly, for $x$ odd, take $a=b=x$ and $m=x+1$ to see that
$$
f(x)=\frac{x^2+x}{2x+3},
$$
which cannot be an integer for $x$ odd.
