2020 CAMO Problems/Problem 1

Problem 1

Let $f:\mathbb R_{>0}\to\mathbb R_{>0}$ (meaning $f$ takes positive real numbers to positive real numbers) be a nonconstant function such that for any positive real numbers $x$ and $y$ , $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y).$ Prove that there is a constant $a>1$ such that $f(x)=\frac{a^x-1}{a^x+1}$ for all positive real numbers $x$ .

Solution

Because $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y)$ , we can find that $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ It's obvious that if there exists two real numbers $x$ and $y$ , which satisfies $f(x)=\frac{a^x-1}{a^x+1}$ and $f(y)=\frac{a^y-1}{a^y+1}$

Then, for $f(x+y)$ , $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ , $f(x+y)=\frac{2*a^x*a^y-2}{2*a^x*a^y+2}$

Then, $f(x+y)=\frac{a^x*a^y-1}{a^x*a^y+1}$

The fraction is also satisfies for $f(x+y)$

Then, we can solve this problem using mathematical induction

~~Andy666

Solution (2)

Let $P(x_0,y_0)$ denote a substitution of $(x_0,y_0)$ for $(x,y)$ and $g$ be the inverse of $f$ when it exists.

By $P(x,x)$ we get $f(x)<1$ so the domain $D$ of $g$ (x) must be in the $0<x<1$ interval

$P(g(x),g(y)); g(x)+g(y)=g(\frac{x+y}{1+xy})$ (*) from here,

Taking $\frac{\partial}{\partial x} and \frac{\partial}{\partial y};$

$g^\prime(x)=\frac{1(1+xy)-y(x+y)}{(1+xy)^2}=\frac{1-y^2}{(1+xy)^2}, g^\prime(y)=\frac{1(1+xy)-x(x+y)}{(1+xy)^2}=\frac{1-x^2}{(1+xy)^2}$

$\frac{g^\prime(x)}{g^\prime(y)}=\frac{1-y^2}{1-x^2}$ so let $g^\prime(x)(1-x^2)=g^\prime(y)(1-y^2)=k$ for some real constant $k>0$ .

$g(x)=\int{\frac{k}{1-x^2}}dx=kartanh(x)+c$

by substitution into (*); $kartanh(x)+kartanh(y)+2c=kartanh(\frac{x+y}{1+xy})+c$ we know that $artanh(x)+artanh(y)=artanh(\frac{x+y}{1+xy})$ so $\frac{c}{k}=0$ so $c=0,k\neq0;g(x)=kartanh(x)$

so

where $a=e^{\frac{1}{k}}>1$

-Shushninja

2020 CAMO (Problems • Resources)
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All CAMO Problems and Solutions

2020 CJMO (Problems • Resources)
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All CJMO Problems and Solutions

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2020 CAMO Problems/Problem 1

Contents

Problem 1

Solution

Solution (2)

See also