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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?

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  • $\begingroup$ 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. $\endgroup$ – Ross Millikan May 30 at 3:23
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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.

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  • $\begingroup$ What do you mean by "Set $y:=y+x$"? $\endgroup$ – JiK May 30 at 12:18
  • $\begingroup$ @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. $\endgroup$ – auscrypt May 30 at 12:22
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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 $x\in\mathbb{Z}^+$.

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

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

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

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

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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 $$\tag2 x,y\in S\land x<y\implies (y-x)x\in S.$$ 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 $x\in\Bbb Q_+$.

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