## Problem 3

Find all $a \in \mathbb{R}$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$, such that

(i) $f(f(x))=f(x)+x$, for all $x \in \mathbb{R}$,

(ii) $f(f(x)-x)=f(x)+$ ax, for all $x \in \mathbb{R}$.


Solution 3 Answer: $a=\frac{1 \pm \sqrt{5}}{2}$.

From (i) we get $f(f(f(x))-f(x))=f(x)$. On the other hand (ii) gives

$$
f(f(f(x))-f(x))=f(f(x))+a f(x)
$$

Thus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\frac{1}{a} x$, since $a=0$ obviously does not give a solution.

We now need to check whether (i) and (ii) hold for this function for some values of $a$ and all real $x$. We have

$$
f(f(x))=-\frac{1}{a} f(x)=\frac{1}{a^{2}} x, \text { and } f(x)+x=-\frac{1}{a} x+x=\frac{a-1}{a} x
$$

Thus (i) will hold for all real $x$ iff $\frac{1}{a^{2}}=\frac{a-1}{a}$, i.e. iff $a=\frac{1 \pm \sqrt{5}}{2}$. For these values of $a$ we have

$$
f(f(x)-x)=-\frac{1}{a}(f(x)-x)=-\frac{1}{a}\left(-\frac{1}{a} x-x\right)=\left(\frac{1}{a^{2}}+\frac{1}{a}\right) x=\frac{a+1}{a^{2}} x=x
$$

and

$$
f(x)+a x=-\frac{1}{a} x+a x=\frac{a^{2}-1}{a} x=x
$$

so that for these two values of $a$ both (i) and (ii) hold for all real $x$. Thus the values of $a$ such that there exists a function $f$ with the desired properties are $a=\frac{1 \pm \sqrt{5}}{2}$.

