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Find all functions $f$: $\mathbb{R}^{+}\rightarrow \mathbb{R}$ such that

$$f\left ( \frac{x}{y} \right )= f(x)+f(y)-f(x)f(y)$$

for all $x,y\in\mathbb{R}^{+}$. Here, $\mathbb{R}^{+}$, denotes the set of all positive real numbers.

I really couldn't solve it. Any help?

This question from IMO Competition.

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    $\begingroup$ Have you tried basic tricks like putting in zero for $x$ or setting $x$ and $y$ equal? $\endgroup$ – Zach Effman Jun 27 '15 at 2:40
  • $\begingroup$ @ZachEffman The domain is $\Bbb Z^+$, so you cannot put zero for $x$. $\endgroup$ – user26486 Jun 27 '15 at 2:40
  • $\begingroup$ Probably we should choose y=1 and then x=y. $\endgroup$ – user72012 Jun 27 '15 at 2:42
  • $\begingroup$ From which IMO is this problem? Which year? $\endgroup$ – user26486 Jun 27 '15 at 3:01
  • $\begingroup$ 2015 day 1, first problem $\endgroup$ – HANSE Jun 27 '15 at 3:13
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Let $P(x,y)$ be the property that $f(\tfrac{x}{y}) = f(x)+f(y)-f(x)f(y)$.

$P(1,1)$ yields $f(1) = 2f(1)-f(1)^2$, and so, either $f(1) = 1$ or $f(1) = 0$.

If $f(1) = 1$, then $P(x,1)$ yields $f(x) = f(x)+f(1)-f(x)f(1)$, which simplifies to $f(x) = 1$ for all $x \in \mathbb{R}^+$.

Now, suppose instead $f(1) = 0$.

$P(x,x)$ yields $f(1) = 2f(x)-f(x)^2$, which gives us that $2f(x)-f(x)^2 = 0$ for all $x \in \mathbb{R}^+$.

$P(1,x)$ yields $f(\tfrac{1}{x}) = f(1)+f(x)-f(1)f(x)$, which simplifies to $f(\tfrac{1}{x}) = f(x)$ for all $x \in \mathbb{R}^+$.

Finally, $P(\sqrt{x},\tfrac{1}{\sqrt{x}})$ yields $f(x) = f(\sqrt{x})+f(\tfrac{1}{\sqrt{x}})-f(\sqrt{x})f(\tfrac{1}{\sqrt{x}}) = 2f(\sqrt{x})-f(\sqrt{x})^2 = 0$ for all $x \in \mathbb{R}^+$.

Hence, the only solutions are $f(x) = 0$ for all $x \in \mathbb{R}^+$ and $f(x) = 1$ for all $x \in \mathbb{R}^+$.

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None of these functions are very interesting. Let $x=y$. Then $$f(1)=2f(x)-f(x)^2.$$ Solving this quadratic, we have that $$f(x)=1\pm\sqrt{1-f(1)}.$$ Thus, $f$, whatever it is, is constant. To find the constant, $$f(1)=1\pm\sqrt{1-f(1)}$$ $$f(1)^2-2f(1)+1=1-f(1)$$ $$f(1)^2-f(1)=0$$ $$f(1)=0\text{ or }f(1)=1.$$ Thus, $f(x)=0$ and $f(x)=1$ are the only two continuous solutions.

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  • $\begingroup$ No, $f(x)$ may jump between $1+\sqrt{1-f(1)}$ and $1-\sqrt{1-f(1)}$ at different points. $\endgroup$ – user26486 Jun 27 '15 at 2:56
  • $\begingroup$ Ok true. I was focused on continuous functions. $\endgroup$ – Alex S Jun 27 '15 at 2:57
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Small Hint :Let $y=1$. $f(x)=f(x)+f(1)-f(x)f(1)$. So if $f$ does not vanish at $1$,then $f$ is the constant function $1$. Edited later: If $f$ vanishes at $1$, then pluging in $x=y$, you can get $2f(x)=f(x)^{2}$.

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There are two approaches to problems like this that I know. One is to recognize some $f$ that meets the requirement and then prove that is the only one. I don't see one offhand, but maybe somebody will. The other is to try special values for the variables and see what you learn. For example, let $y=1$, then $f(x)=f(x)+f(1)-f(x)f(1)$ or $0=f(1)(1-f(x))$

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