# Complicated rational number functional equation

Let $$\mathbb{Q}^+$$ denote the set of positive rational numbers. Let $$f : \mathbb{Q}^+ \to \mathbb{Q}^+$$ be a function such that $$f \left( x + \frac{y}{x} \right) = f(x) + \frac{f(y)}{f(x)} + 2y$$ for all $$x,$$ $$y \in \mathbb{Q}^+.$$

Find all possible values of $$f \left( \frac{1}{3} \right).$$

If I substitute in $$x=y$$, then I get $$f(x+1)-f(x)=2x+1$$. This suggests that $$f(x)=x^2$$ works, and one possible value of $$f(1/3)$$ is $$1/9$$. Did I miss anything?

• Your try of $x=y$ is a good start and it does suggest that $f(x)=x^2$ works, but you should plug that into the defining equation and see if it does. Then you have to think about whether there are other possibilities. – Ross Millikan May 30 at 3:23

Yes.

But what you didn't do was prove that you had all possible values. If somebody told you to solve $$x^2=1$$, you wouldn't say $$1$$ and leave it. You would prove that $$1$$ is the only solution, or find more solutions and prove that you'd found all of them.

A similar things applies here. The question asks for you to find all possible values. How do you know you have all of them?

Here's a quick sketch of how to prove that you have all solutions: Set $$y:=y+x$$ and subtract the original equation to get $$2x+2\frac{y}{x}+1=\frac{f(y+x)-f(y)}{f(x)}+2x$$Rearrange and simplify:

$$2\frac{y}{x}f(x)+f(x)=f(y+x)-f(y)$$ Repeat the same trick: set $$y:=y+1$$ and subtract that last equation to get $$2\frac{f(x)}{x}=2y+2x+1-2y-1$$ so $$f(x)=x^2$$ as required. Just verify this works by substituting back into the original equation.

• What do you mean by "Set $y:=y+x$"? – JiK May 30 at 12:18
• @JiK If you like you can consider it as splitting $y$ up into two variables, $x+z= y$, and then relabelling $z$ to $y$ as a dummy variable. – auscrypt May 30 at 12:22

From $$f(x+1)-f(x)=2x+1$$, we can deduce that $$f(x)=x^2+f(1)-1$$ for all $$x\in\mathbb{Z}^+$$.

Note that we also have $$f(1+\frac x1)=f(1)+\frac{f(x)}{f(1)}+2x$$.

Therefore, $$f(x+1)=f(x)+2x+1=\dfrac{f(x)}{f(1)}+2x+f(1)$$.

Hence, $$f(x)\left(1-\dfrac{1}{f(1)}\right)=f(1)-1$$ for $$x\in\mathbb{Q}^+$$.

As $$f$$ is not constantly zero, $$f(1)=1$$. So, $$f(x)=x^2$$ for $$x\in\mathbb{Z}^+$$.

$$f(3+\frac{1}{3})=f(3)+\frac{f(1)}{f(3)}+2(1)=3^2+\frac{1}{3^2}+2=11+\frac19$$

$$f(2+\frac{1}{3})=f(3+\frac{1}{3})-2(2+\frac13)-1=5+\frac49$$

$$f(1+\frac{1}{3})=f(2+\frac{1}{3})-2(1+\frac13)-1=1+\frac79$$

$$f(\frac{1}{3})=f(1+\frac{1}{3})-2(\frac13)-1=\frac19$$

We have $$\tag{x=y=1} f(2)=f(1)+3$$ $$\tag{x=1,y=2} f(3)=f(1)+\frac{f(2)}{f(1)}+4=f(1)+5+\frac3{f(1)}$$ $$\tag{x=y=2} f(3)=f(2)+5=f(1)+8$$ hence $$f(1)=1$$. Let $$S=\{\,x\in\Bbb Q_+\mid f(x)=x^2\,\}$$. As just seen, $$1\in S$$. From the functional equation we see that if $$x$$ and one of $$y,x+\frac yx$$ are $$\in S$$, then so is the third. In other words, $$\tag1x,y\in S\implies x+\frac yx\in S$$ and $$\tag2 x,y\in S\land x In particular, using $$(1)$$ with $$y=x$$ and $$(2)$$ with $$x=1$$, we find that for $$x\in \Bbb Q_+$$, $$\tag3x\in S\iff x+1\in S$$ and so by induction $$\Bbb Z_+\subseteq S$$.

Let $$x=\frac ab\in\Bbb Q_+$$. From $$(1)$$, $$b+x\in S$$. Then by applying $$(3)$$ $$b$$ times, $$x\in S$$. In other words, $$S=\Bbb Q_+$$ and $$f(x)=x^2$$ for all $$x\in\Bbb Q_+$$.