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

Epicskills (talk | contribs) (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> | ||

+ | |||

+ | ==Solution== | ||

+ | |||

+ | 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>. | ||

+ | |||

+ | Let <math>f(0)=n</math>, so <math>f(n^{2})=0</math>. | ||

+ | |||

+ | Let <math>y=n^2</math>, so <math>f(x+n^2)=f(xn^2)-n</math>. | ||

+ | |||

+ | 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. | ||

+ | |||

+ | This means that the only possible values of <math>n</math> is -1,0, and 1. | ||

+ | |||

+ | Go through the cases: | ||

+ | |||

+ | <math>n=-1</math> | ||

+ | |||

+ | <math>f(x+1)=f(x)+1</math> | ||

+ | |||

+ | <math>f(1)=0</math> (Based on <math>f(n^2)=0</math>) | ||

+ | |||

+ | <math>f(2)=1</math> | ||

+ | |||

+ | <math>f(3)=2</math> | ||

+ | |||

+ | ... | ||

+ | |||

+ | <math>f(x)=x-1</math> | ||

+ | |||

+ | |||

+ | <math>n=0</math> | ||

+ | |||

+ | <math>f(x)=0</math> | ||

+ | |||

+ | |||

+ | <math>n=1</math> | ||

+ | |||

+ | <math>f(x+1)=f(x)-1</math> | ||

+ | |||

+ | <math>f(1)=0</math> (Based on <math>f(n^2)=0</math>) | ||

+ | |||

+ | <math>f(2)=-1</math> | ||

+ | |||

+ | <math>f(3)=-2</math> | ||

+ | |||

+ | ... | ||

+ | |||

+ | <math>f(x)=1-x</math> | ||

+ | |||

+ | |||

+ | 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 be the set of real numbers , determine all functions such that for any real numbers and =

## Solution

Let , so the equation becomes . Notice that if , , so , meaning that there exists at least 1 real solution to .

Let , so .

Let , so .

If , or , then , so . The only way n can not equal 0 is if there is no solution to , so if does not equal 0.

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

Go through the cases:

(Based on )

...

(Based on )

...

So the only solutions are , , and .