China Girls' Mathematical Olympiad

By symmetry, we may assume that $a\ge c$. We claim that $a=c$. Assume on the contrary that $a>c$. Because $a>c>0$, the ranges of $f$ and $g$ are both $\mathbf{R}$. There is a $x_0$ such that $f(x_0)=a{x}_0+b$ is an integer. Hence $g(x_0)=c{x}_0+d$ is also an integer. But then, $$f\left(x_0+\frac{1}{a}\right)=a{x}_0+b+1$$ and $$g\left(x_0+\frac{1}{a}\right)=c{x}_0+d+\frac{c}{a}.$$ But this is impossible because we cannot have two integers $g(x_0)$ and $g(x_1)$ that have positive difference $\frac{c}{a}$ which is less than 1. Therefore, we can write $f(x)=ax+b$ and $g(x)=ax+d$ for some real numbers $a,b,d$ with $a>0$. Then $b-d=f(x_0)-g(x_0)$ must be an integer, that is, $$f(x)-g(x)=b-d$$ is an integer. □