Difference between revisions of "2017 IMO Problems/Problem 2"

(Created page with "Let <math>\mathbb{R}</math> be the set of real numbers , determine all functions <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for any real numbers <math>x</math>...")
 
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Let <math>\mathbb{R}</math> be the set of real numbers , determine all functions  
 
Let <math>\mathbb{R}</math> be the set of real numbers , determine all functions  
 
<math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for any real numbers <math>x</math> and <math>y</math> <math>{f(f(x)f(y)) + f(x+y)}</math> =<math>f(xy)</math>
 
<math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for any real numbers <math>x</math> and <math>y</math> <math>{f(f(x)f(y)) + f(x+y)}</math> =<math>f(xy)</math>
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==Solution==
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Let <math>y=x</math>, so the equation becomes <math>f(f(x)^{2})+f(2x)=f(x^{2})</math>. Notice that if <math>x=0</math>, <math>2x=x^2</math>, so <math>f(f(0)^{2})=0</math>, meaning that there exists at least 1 real solution to <math>f(x)=0</math>.
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Let <math>f(0)=n</math>, so <math>f(n^{2})=0</math>.
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Let <math>y=n^2</math>, so <math>f(x+n^2)=f(xn^2)-n</math>.
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If <math>x+n^2=xn^2</math>, or <math>x=\frac{n^2}{n^2-1}</math>, then <math>f(x+n^2)=f(xn^2)</math>, so <math>n=0</math>. The only way n can not equal 0 is if there is no solution to <math>x+n^2=xn^2</math>, so <math>n^2=1</math> if <math>n</math> does not equal 0.
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This means that the only possible values of <math>n</math> is -1,0, and 1.
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Go through the cases:
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<math>n=-1</math>
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<math>f(x+1)=f(x)+1</math>
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<math>f(1)=0</math> (Based on <math>f(n^2)=0</math>)
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<math>f(2)=1</math>
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<math>f(3)=2</math>
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...
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<math>f(x)=x-1</math>
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<math>n=0</math>
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<math>f(x)=0</math>
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<math>n=1</math>
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<math>f(x+1)=f(x)-1</math>
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<math>f(1)=0</math> (Based on <math>f(n^2)=0</math>)
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<math>f(2)=-1</math>
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<math>f(3)=-2</math>
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...
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<math>f(x)=1-x</math>
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So the only solutions are <math>f(x)=x-1</math>, <math>f(x)=0</math>, and <math>f(x)=1-x</math>.

Revision as of 11:35, 9 December 2020

Let $\mathbb{R}$ be the set of real numbers , determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any real numbers $x$ and $y$ ${f(f(x)f(y)) + f(x+y)}$ =$f(xy)$

Solution

Let $y=x$, so the equation becomes $f(f(x)^{2})+f(2x)=f(x^{2})$. Notice that if $x=0$, $2x=x^2$, so $f(f(0)^{2})=0$, meaning that there exists at least 1 real solution to $f(x)=0$.

Let $f(0)=n$, so $f(n^{2})=0$.

Let $y=n^2$, so $f(x+n^2)=f(xn^2)-n$.

If $x+n^2=xn^2$, or $x=\frac{n^2}{n^2-1}$, then $f(x+n^2)=f(xn^2)$, so $n=0$. The only way n can not equal 0 is if there is no solution to $x+n^2=xn^2$, so $n^2=1$ if $n$ does not equal 0.

This means that the only possible values of $n$ is -1,0, and 1.

Go through the cases:

$n=-1$

$f(x+1)=f(x)+1$

$f(1)=0$ (Based on $f(n^2)=0$)

$f(2)=1$

$f(3)=2$

...

$f(x)=x-1$


$n=0$

$f(x)=0$


$n=1$

$f(x+1)=f(x)-1$

$f(1)=0$ (Based on $f(n^2)=0$)

$f(2)=-1$

$f(3)=-2$

...

$f(x)=1-x$


So the only solutions are $f(x)=x-1$, $f(x)=0$, and $f(x)=1-x$.

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