APMO 2016

The identity function $f(x)=x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x=y=1$ into (3) we get $2f(f(z)+1)=(z+1)f(2)$ for all $z\in {\mathbb{R}}^{+}$. Hence, $f$ is not bounded above.

Lemma. Let $a,b,c$ be positive real numbers. If $c$ is greater than $1,a/b$ and $b/a$, then the system of linear equations $$cu+v=au+cv=b$$ has a positive real solution $u,v$. Proof. The solution is $$u=\frac{ca-b}{c^2-1}v=\frac{cb-a}{c^2-1}$$ The numbers $u$ and $v$ are positive if the conditions on $c$ above are satisfied.

We will now prove that $$f(a)+f(b)=f(c)+f(d)\text{ for all }a,b,c,d\in {\mathbb{R}}^{+}\text{ with }a+b=c+d.$$ Consider $a,b,c,d\in {\mathbb{R}}^{+}$ such that $a+b=c+d$. Since $f$ is not bounded above, we can choose a positive number $e$ such that $f(e)$ is greater than $1,a/b,b/a,c/d$ and $d/c$. Using the above lemma, we can find $u,v,w,t\in {\mathbb{R}}^{+}$ satisfying $$\begin{array}{ll}f(e)u+v=a,& u+f(e)v=b\\ f(e)w+t=c,& w+f(e)t=d.\end{array}$$ Note that $u+v=w+t$ since $(u+v)(f(e)+1)=a+b$ and $(w+t)(f(e)+1)=c+d$. Plugging $x=u,y=v$ and $z=e$ into (3) yields $f(a)+f(b)=(e+1)f(u+v)$. Similarly, we have $f(c)+f(d)=(e+1)f(w+t)$. The claim follows immediately. We then have $$yf(x)=f(xf(y))\text{ for all }x,y\in {\mathbb{R}}^{+}$$ since by (3) and (4), $$(y+1)f(x)=f\left(\frac{x}{2}f(y)+\frac{x}{2}\right)+f\left(\frac{x}{2}f(y)+\frac{x}{2}\right)=f(xf(y))+f(x).$$ Now, let $a=f(1/f(1))$. Plugging $x=1$ and $y=1/f(1)$ into (5) yields $f(a)=1$. Hence $a=af(a)$ and $f(af(a))=f(a)=1$. Since $af(a)=f(af(a))$ by (5), we have $f(1)=a=1$. It follows from (5) that $$f(f(y))=y\text{ for all }y\in {\mathbb{R}}^{+}.$$ Using (4) we have for all $x,y\in {\mathbb{R}}^{+}$ that $$\begin{array}{rl}& f(x+y)+f(1)=f(x)+f(y+1),\text{ and }\\ & f(y+1)+f(1)=f(y)+f(2).\end{array}$$ Therefore $$f(x+y)=f(x)+f(y)+b\text{ for all }x,y\in {\mathbb{R}}^{+}$$ where $b=f(2)-2f(1)=f(2)-2$. Using (5), (7) and (6), we get $$4+2b=2f(2)=f(2f(2))=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b.$$ This shows that $b=0$ and thus $$f(x+y)=f(x)+f(y)\text{ for all }x,y\in {\mathbb{R}}^{+}.$$ In particular, $f$ is strictly increasing. We conclude as follows. Take any positive real number $x$. If $f(x)>x$, then $f(f(x))>f(x)>x=f(f(x))$, a contradiction. Similarly, it is not possible that $f(x)<x$. This shows that $f(x)=x$ for all positive real numbers $x$.