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Difference between revisions of "2017 IMO Problems/Problem 2"
 
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==Problem==
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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  
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<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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<math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for any real numbers <math>x</math> and <math>y</math>
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<cmath>f(f(x)f(y)) + f(x+y)=f(xy)</cmath>
  
 
==Solution==
 
==Solution==
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Looking at the equation one can deduce that the functions that will work will be linear.  That is, a polynomial of at most a degree of 1.
  
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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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Thus, <math>f</math> is in the form <math>f(x)=mx+b</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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Therefore,
  
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<math>n=-1</math>
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<math>f((mx+b)(my+b))+m(x+y)+b=mxy+b</math>
  
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<math>f(x+1)=f(x)+1</math>
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<math>f(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b</math>
  
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<math>f(1)=0</math> (Based on <math>f(n^2)=0</math>)
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<math>m(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b</math>
  
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<math>f(2)=1</math>
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<math>m^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy</math>
  
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<math>f(3)=2</math>
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<math>m(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0</math>
  
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...
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Therefore,
  
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<math>f(x)=x-1</math>
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<math>m(m^2-1)=0</math> [Equation 1]
  
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<math>m(bm+1)=0</math> [Equation 2]
  
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<math>n=0</math>
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<math>b(bm+1)</math> [Equation 3]
  
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<math>f(x)=0</math>
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From [Equation 1] we have, <math>m=0,\pm 1</math>
  
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From [Equation 2] we have, <math>m=0, bm=-1\pm 1</math>
  
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<math>n=1</math>
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From [Equation 3] we have, <math>b=0, bm=-1\pm 1</math>
  
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<math>f(x+1)=f(x)-1</math>
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When <math>m=0</math>, <math>b=0</math>, then <math>f(x)=0</math>
  
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<math>f(1)=0</math> (Based on <math>f(n^2)=0</math>)
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When <math>bm=-1</math>, <math>b=\frac{-1}{m}</math>, then since <math>m=\pm 1</math>, then <math>b=\mp 1</math>
  
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<math>f(2)=-1</math>
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When <math>bm=-1</math>, <math>b=\frac{-1}{m}</math>, then since <math>m=\pm 1</math>, then <math>b=\mp 1</math> then <math>f(x)=\pm x \mp1</math> which gives these two functions:
  
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<math>f(3)=-2</math>
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<math>f(x)=x-1</math> and <math>f(x)=1-x</math>, which with <math>f(x)=0</math> provide all the three functions for this problem.
  
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...
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~Tomas Diaz. orders@tomasdiaz.com
  
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<math>f(x)=1-x</math>
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{{alternate solutions}}
  
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==See Also==
  
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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>.
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{{IMO box|year=2017|num-b=1|num-a=3}}

Latest revision as of 10:15, 20 November 2023

Problem

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

Looking at the equation one can deduce that the functions that will work will be linear. That is, a polynomial of at most a degree of 1.

Thus, $f$ is in the form $f(x)=mx+b$

Therefore,

$f((mx+b)(my+b))+m(x+y)+b=mxy+b$

$f(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b$

$m(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b$

$m^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy$

$m(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0$

Therefore,

$m(m^2-1)=0$ [Equation 1]

$m(bm+1)=0$ [Equation 2]

$b(bm+1)$ [Equation 3]

From [Equation 1] we have, $m=0,\pm 1$

From [Equation 2] we have, $m=0, bm=-1\pm 1$

From [Equation 3] we have, $b=0, bm=-1\pm 1$

When $m=0$, $b=0$, then $f(x)=0$

When $bm=-1$, $b=\frac{-1}{m}$, then since $m=\pm 1$, then $b=\mp 1$

When $bm=-1$, $b=\frac{-1}{m}$, then since $m=\pm 1$, then $b=\mp 1$ then $f(x)=\pm x \mp1$ which gives these two functions:

$f(x)=x-1$ and $f(x)=1-x$, which with $f(x)=0$ provide all the three functions for this problem.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2017 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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