2016 APMO Problems/Problem 5

Problem

Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),$ for all positive real numbers $x, y, z$ .

Solution

We claim that $f(x)=x$ is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let $P(x.y,z)$ be the assertion to the Functional Equation.

Claim 1: $f$ is injective.

Proof: Assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$ . Now, from $P(x,y,a)$ and $P(x,y,b)$ we have:

$(a+1)f(x+y)=f(xf(a)+y)+f(yf(a)+x)$ $(b+1)f(x+y)=f(xf(b)+y)+f(yf(b)+x)$

Now comparing, we have $a=b$ as desired. $\square$

This gives us the power to compute $f(1)$ . From $P(1,1,1)$ we get $f(f(1)+1)=f(2)$ and injectivity gives $f(1)=1$ . Showing that $f$ is unbounded above is also easy as we can fix $(x,y)$ and let $z$ blow up to $\infty$ in the original Functional equation..

Claim 2: $f$ is surjective.

Proof: $P(x,x,z)$ gives $(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))$

This gives that $\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)$ . Putting $x=\frac{1}{2}$ , we get $\text{Im}(f) \in \left(\frac{1}{2},\infty\right)$ . By induction, surjectivity is proved as $\lim_{m \to \infty}\frac{1}{2^m}=0$ and we are essentially done. $\square$

Now, we have claim that if $a+b=c+d$ for some $a,b,c,d \in \mathbb{R}^+$ , then $f(a)+f(b)=f(c)+f(d)$ . This can be achieved by putting $a=xf(z)+y$ , $b=yf(z)+x$ , $c=x'f(z)+y'$ and $d=y'f(z)+x'$ . Let us calculate $f(a)+f(b)$ now. e $f(a)+f(b)=(z+1)f(x+y)=(z+1)f\left(\frac{a+b}{f(z)+1}\right)=(z+1)f\left(\frac{a'+b'}{f(z)+1}\right)=(z+1)f(x'+y')=f(a')+f(b').$

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2016 APMO Problems/Problem 5

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