## Vietnam 2012- Surjective function

For discussing Olympiad Level Algebra (and Inequality) problems
*Mahi*
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### Vietnam 2012- Surjective function

Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$
1)$f$ is surjective
2)For $x>y$, $f(x)>f(y)$
3)$f(f(x))=f(x)+12x$
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Nur Muhammad Shafiullah | Mahi

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### Re: Vietnam 2012- Surjective function

$f(f(0))=f(0)$ and strictly increasing property yields $f(0)=0$.
let $a_n=f^{n}(x)$ and let $a_0=x$
then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$
$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$
but $f(0)=0$ implies $q=0$
so $f(x)=-3x$ which is surjective and strictly increasing.

*Mahi*
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### Re: Vietnam 2012- Surjective function

zadid xcalibured wrote: so $f(x)=-3x$ which is surjective and strictly increasing.
$f(x) = -3x$ is strictly decreasing.
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Nur Muhammad Shafiullah | Mahi

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### Re: Vietnam 2012- Surjective function

The mistake in Zadid vai's solution was perhaps to write $a_n=ap^n+bq^n$. As far as I know the right form should be $a_n=pa^n+qb^n$ and this yields a valid solution $f(x)=4x$
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

There are two cases.I missed it.One case gives $a=0$ and $b=x$ another cases gives $b=0$ and $a=x$.One case gives no solution while the other case gives $f(x)=4x$