Solutions

a) Notice that

$$\begin{align*} f(x_1+\cdots+x_n) &= f((x_1+\cdots+x_{n-1})+x_n) \\ &= f(x_1+\cdots+x_{n-1})+f(x_n) \\ &\vdots \\ &= f(x_1)+\cdots+f(x_n). \end{align*}$$

b) For $x\in\mathbb{N}^*$, we can write

$$\begin{align*} f(x) &= f(\underbrace{1+\cdots+1}_{x\text{ times}}) \\ &= \underbrace{f(1)+\cdots+f(1)}_{x\text{ times}} \\ &= f(1)x. \end{align*}$$

So $c=f(1)$. For $x=0$, we have that

$$\begin{align*} f(0)&=f(0+0) \\ &=f(0)+f(0), \end{align*}$$

so $f(0)=0$ and we can set $c=0$. If $x$ is a negative integer, we can write

$$\begin{align*} f(x) &= f(\underbrace{-1+\cdots+(-1)}_{x\text{ times}}) \\ &= \underbrace{f(-1)+\cdots+f(-1)}_{x\text{ times}} \\ &= f(-1)x. \end{align*}$$

So $c=f(-1)$. For the reciprocal of a positive integer $x$ (the result is similar for a negative integer),

$$\begin{align*} f\left(\frac{1}{x}\right) = f\left(\frac{x}{x^2}\right) = f\left(\underbrace{\frac{1}{x^2}+\cdots+\frac{1}{x^2}}_{x\text{ times}}\right) = f\left(\frac{1}{x^2}\right)x. \end{align*}$$

So $c=f\left(\frac{1}{x^2}\right)$. Finally, for $x=\frac{m}{n}$ where $m,n\in\mathbb{Z}$, we have

$$\begin{align*} f\left(\frac{m}{n}\right) &= m f\left(\frac{1}{n}\right) \\ &= n f\left(\frac{1}{n^2}\right)x. \end{align*}$$

So

$$c=n f\left(\frac{1}{n^2}\right).$$