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

Latest revision as of 09:15, 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.

@@ Line 4: / Line 4: @@
 <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for any real numbers <math>x</math> and <math>y</math>
-<math></math>{f(f(x)f(y)) + f(x+y)}<math> =</math>f(xy)<math></math>
+<cmath>f(f(x)f(y)) + f(x+y)=f(xy)</cmath>
 ==Solution==
-{{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, <math>f</math> is in the form <math>f(x)=mx+b</math>
+Therefore,
+<math>f((mx+b)(my+b))+m(x+y)+b=mxy+b</math>
+<math>f(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b</math>
+<math>m(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b</math>
+<math>m^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy</math>
+<math>m(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0</math>
+Therefore,
+<math>m(m^2-1)=0</math> [Equation 1]
+<math>m(bm+1)=0</math> [Equation 2]
+<math>b(bm+1)</math> [Equation 3]
+From [Equation 1] we have, <math>m=0,\pm 1</math>
+From [Equation 2] we have, <math>m=0, bm=-1\pm 1</math>
+From [Equation 3] we have, <math>b=0, bm=-1\pm 1</math>
+When <math>m=0</math>, <math>b=0</math>, then <math>f(x)=0</math>
+When <math>bm=-1</math>, <math>b=\frac{-1}{m}</math>, then since <math>m=\pm 1</math>, then <math>b=\mp 1</math>
+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:
+<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.
+~Tomas Diaz. orders@tomasdiaz.com
+{{alternate solutions}}
 ==See Also==
 {{IMO box|year=2017|num-b=1|num-a=3}}

2017 IMO (Problems) • Resources
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Difference between revisions of "2017 IMO Problems/Problem 2"

Latest revision as of 09:15, 20 November 2023

Problem

Solution

See Also