Since $f$ and $g$ are linear functions, they are bijective. Suppose $g(a)=0$, and let $f(a)=b$ with $b$ an integer. Then suppose that $g(c)=1$. Then $f(c)\geq b+1$ because it's increasing and it has to be an integer. Also $f(c)\leq b+1$ because otherwise, there is some value between $a$ and $c$ so that the value there is $b+1$, impossible. Thus $f(c)=b+1$. Then $g$ and $f$ have the same slope, $\frac{1}{c-a}$, so their difference is constant, which is $b$, an integer.

Let $f(x)=ax+b$ and $g(x)=cx+d$. Consider $r$ where both are integers, i.e. \[ar+b,cr+d\in\mathbb{Z}.\]Then \[f\left(r+\frac1a\right)=ar+b+1\in\mathbb{Z}\implies g\left(r+\frac1a\right)=cr+d+\frac{c}{a}\in\mathbb{Z},\]so $a|c$. Similarly, $c|a$, so $a=c$, and we're clearly done.

Let $f(x)=ax+b$ and $g(x)=cx+d$ $f$ and $g$ are strictly increasing functions so $a>0$ and $c>0$ are $f\left(-\frac{b}{a}\right)=0$ so $g\left(-\frac{b}{a}\right)=d-\frac{bc}{a}\in\mathbb{Z}$ $f\left(-\frac{b+1}{a}\right)=-1$ so $g\left(-\frac{b+1}{a}\right)=d-\frac{bc+c}{a}\in\mathbb{Z}$ So $\frac{c}{a}=\left(d-\frac{bc}{a}\right)-\left(d-\frac{bc+c}{a}\right)\in\mathbb{Z}$ so $a|c$ Similarly $c|a$ so $a=c$ or $a+c=0$ But $a>0$ and $c>0$ so $a=c$ Now $d-\frac{bc}{a}$ is an integer so $d-b$ is an integer so $b-d$ is an integer. Hence $f(x)-g(x)$ is an integer for all reals $x$, as required.

let $f(x)=ax+b , g(x)=cx+d $,where $a>0 , c>0 $ since$ f,g$ are strictly increasing functions. let $F=\{\frac{k-b}{a} :k \in \mathbb{Z} \} ,G=\{\frac{k-d}{c} :k \in \mathbb{Z} \}$, if $x\in F$ ,there exists $k\in \mathbb{Z}$ such that $x=\frac{k-b}{a}$ so $f(x)=k \in \mathbb{Z}$ If$ f(x) \in \mathbb{Z}$,$x=\frac{f(x)-b}{a} \in F$ $\therefore f(x) \in \mathbb{Z} \iff x \in F$. similarly , $g(x) \in \mathbb{Z} \iff x \in G$ by the condition , $g(x) \in \mathbb{Z} \iff f(x) \in \mathbb{Z}$ $ \therefore x \in F \iff x\in G$,so we can get $F=G$. let$ p_n=\frac{n-b}{a}, q_n=\frac{n-d}{c}$,\ since $F=G$ , there exists $t \in \mathbb{Z}$ such that $p_0=q_t.....(1)$, obviously, $\{p_n\}$ and $\{q_n\} $are strictly increasing sequences. thus, $q_t=p_0< p_1$ and $p_0=q_t<q_{t+1}$ if $q_t< p_1<q_{t+1}$,then there is no term in$ \{q_n\}$ equal to$p_1 $ since $ \{q_n\}$ is strictly increasing,which contradicts $F=G$ similarly, it is impossible that $p_0=q_t<q_{t+1}<p_1$. As a result, $q_{t+1}=p_1.....(2)$ $(1)-(2):\frac{1}{a}=\frac{1}{c}$ so $a=c$ from (1):$\frac{-b}{a}=\frac{t-d}{c}$ so $b-d=-t \in \mathbb{Z}$ Hence, $f(x)-g(x)=b-d=-t \in \mathbb{Z} ,\forall x \in \mathbb{R}$

Let $f(x)=ax+b$ and $g(x)=cx+d$ for any real $a,b,c,d$ with $a,c>0.$ WLOG $a\geq c.$ Claim: $a=c.$ Proof.Suppose not. The range of both $f$ and $g$ is the set of reals since $a>c>0.$ There is an $y$ s.t. $$f(y)=ay+b \in \mathbb{Z} \iff g(y)=cy+d \in \mathbb{Z}.$$Plugging in $$y+a^{-1} \implies f(y+a^{-1})= ay+b+1 \implies g(y+a^{-1}) =cy+d+ca^{-1},$$a contradiction since two integers cannot have a positive difference of $ca^{-1}<1.$ $\blacksquare$ With this it follows that $b-d=f(x)-g(x)\in \mathbb{Z}.$ $\mathbb{Q.E.D.}$