1.3. Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a function from the set $\mathbb{R}$ of real numbers to itself. Find all such functions $f$ satisfying the two properties:
(a) $f(x+f(y))=y+f(x)$ for all $x, y \in \mathbb{R}$,
(b) the set $\left\{\frac{f(x)}{x}: x\right.$ is a nonzero real number $\}$ is finite.
1.3. 1st solution

Let $x \in \mathbb{R}$. By letting $x=y+f(0)$, we obtain
$$
f(f(x))=f(f(y+f(0)))=f(0+f(y))=y+f(0)=x .
$$

Hence for any $t_{1}, t_{2} \in \mathbb{R}, \quad f\left(t_{1}+t_{2}\right)=f\left(t_{1}+f\left(f\left(t_{2}\right)\right)\right)=f\left(t_{1}\right)+f\left(t_{2}\right)$. Next, consider any positive integer $m$ such that $m \neq-f(x)$. We have
$$
\frac{f(m+f(x))}{m+f(x)}=\frac{x+f(m)}{m+f(x)}=\frac{x+m f(1)}{m+f(x)} .
$$

Since the set $\left\{\left.\frac{f(t)}{t} \right\rvert\, t \neq 0\right\}$ is finite, there exist distinct positive integers $m_{1}, m_{2}$ with $m_{1}, m_{2} \neq-f(x)$ such that
$$
\frac{f\left(m_{1}+f(x)\right)}{m_{1}+f(x)}=\frac{f\left(m_{2}+f(x)\right)}{m_{2}+f(x)} .
$$

Hence $\frac{x+m_{1} f(1)}{m_{1}+f(x)}=\frac{x+m_{2} f(1)}{m_{2}+f(x)}$. From this, we have $f(x) f(1)=x$.
By letting $x=1$, we obtain $[f(1)]^{2}=1$ so that $f(1)= \pm 1$. Consequently, $f(x)= \pm x$. Also the functions $f(x)=x$ and $f(x)=-x$ clearly satisfy the two given conditions.
2nd solution
(i) First we prove that $f(0)=0$. Putting $x=0=y$, we have $f(f(0))=f(0)$. If $f(0)=a$, then $f(0)=f(f(0))=f(a)$. Thus $a+f(0)=f(0+f(a))=f(f(0))=f(0)$, whence $a=0$.
(ii) Putting $x=0$, we have $f(f(y)=y$ for all $y$.
(iii) We will prove that $f(x)= \pm x$ for all $x$.

Suppose for some $p, f(p)=c p$ for some constant $c \neq \pm 1$. Then $f(p+f(p))=p+f(p)$. Let $q=p+f(p)$. Then $q \neq 0$ and $f(q)=q$. Thus $f(q+f(q))=q+f(q)$ and $f(2 q)=2 q$. Inductively we have $f(n q)=n q$ for any positive integer $n$. Now $f(n q+f(p))=p+f(n q)$. So $f(n q+c p)=p+n q$. Thus $f(n q+c p) /(n q+c p)=1-(c-1) p /(n q+c p)$. Since $c-1 \neq 0$ and there are infinitely many choices for $n$ so that $n q+c p \neq 0$, this gives an infinite number of members in the set $\{f(x) / x\}$ contradicting the second condition. Thus $c= \pm 1$.
(iv) For $f(p)=p$, we will prove that $f(x)=x$ for all $x$.

If $\mathrm{f}(-p)=p$, then $-p=f(f(-p))=f(p)=p$ which is impossible. Thus $f(-p)=-p$. Suppose there exists $r$ such that $f(r)=-r$. Then $f(r+f(p))=p+f(r)$, i.e., $f(r+p)=p-r$. Therefore $f(r+p) /(r+p)=(p-r) /(r+p) \neq \pm 1$. (Note that the denominator is not zero.) (v) From the above we conclude that either $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$. Clearly these functions satisfy the two given conditions. Thus these are the only two functions required.