Let $x$ and $y$ be any two elements in the domain $(Z)$, such that
$f(x+y)=f(x)+f(y)$ $\ldots$ (i)
Differentiating above expression w.r.t ' $y$ ', keeping $x$ constant, we get
$f^{\prime}(x+y)=f^{\prime}(y)$
Let $y=0 \Rightarrow f^{\prime}(x+0)=f^{\prime}(0)$
and Assume $f^{\prime}(0)=K$
$\therefore \quad f^{\prime}(x)=K$
Integrating on both sides, we get
$f(x)=K x+C\{C \rightarrow \text { integration constant }\}$ $\ldots$ (ii)
Now putting $x=y=0$ in Eq. (i), we get $f(0+0)=f(0)+f(0) \Rightarrow f(0)=2 f(0)$ or $f(0)=0$
Let $x=0$, from Eq. (ii)
$f(0)=K \times 0+C \text { or } 0=0+C \text { or } C=0$
$\therefore \quad f(x)=k x$
Case (i) If $k>0$, then $f(x)$ is strictly increasing.
Case (ii) If $k < 0$, then $f(x)$ is strictly decreasing So, function is injective Also $f(x)$ where every element in the codomain is a valid output of the function i.e., range is equal to codomain.
So, function is surjective also.
Therefore, there are two bijective functions.