Difference between revisions of "2014 USAJMO Problems/Problem 3"
(solution is incorrect; needs fixing) |
|||
| Line 12: | Line 12: | ||
Let's say we can find <math>f(x)=x^2, f(y)=0,</math> and <math>x,y\neq 0.</math> Then <cmath>xf(-x)+y^2f(2x)=f(x)^2/x.</cmath> | Let's say we can find <math>f(x)=x^2, f(y)=0,</math> and <math>x,y\neq 0.</math> Then <cmath>xf(-x)+y^2f(2x)=f(x)^2/x.</cmath> | ||
| − | <cmath>y^2f(2x)=x-x^3.</cmath> | + | <cmath>y^2f(2x)=x-x^3.</cmath> (NEEDS FIXING: <math>f(x)^2/x= x^4/x = x^3</math>, so the RHS is <math>0</math> instead of <math>x-x^3</math>.) |
If <math>f(2x)=4x^2,</math> then <math>y^2=\frac{x-x^3}{4x^2}=\frac{1-x^2}{4x},</math> which is only possible when <math>y=0.</math> This contradicts our assumption. Therefore, <math>f(2x)=0.</math> This forces <math>x=\pm 1</math> due to the right side of the equation. Let's consider the possibility <math>f(2)=0, f(1)=1.</math> Substituting <math>(x,y)=(2,1)</math> into the original equation yields <cmath>0=2f(0)+1f(2)=0+f(1)=1,</cmath> which is impossible. So <math>f(2)=f(-2)=4</math> and there are no solutions "combining" <math>f(x)=x^2</math> and <math>f(x)=0.</math> | If <math>f(2x)=4x^2,</math> then <math>y^2=\frac{x-x^3}{4x^2}=\frac{1-x^2}{4x},</math> which is only possible when <math>y=0.</math> This contradicts our assumption. Therefore, <math>f(2x)=0.</math> This forces <math>x=\pm 1</math> due to the right side of the equation. Let's consider the possibility <math>f(2)=0, f(1)=1.</math> Substituting <math>(x,y)=(2,1)</math> into the original equation yields <cmath>0=2f(0)+1f(2)=0+f(1)=1,</cmath> which is impossible. So <math>f(2)=f(-2)=4</math> and there are no solutions "combining" <math>f(x)=x^2</math> and <math>f(x)=0.</math> | ||
Therefore our only solutions are <math>\boxed{f(x)=0}</math> and <math>\boxed{f(x)=x^2.}</math> | Therefore our only solutions are <math>\boxed{f(x)=0}</math> and <math>\boxed{f(x)=x^2.}</math> | ||
Revision as of 14:30, 27 June 2019
Problem
Let 
be the set of integers. Find all functions 
such that ![\[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\]](http://latex.artofproblemsolving.com/9/f/d/9fdb11dbf102dfe0783b1d31466e276420e3913d.png)
for all 
with 
.
Solution
Let's assume 
Substitute 
to get ![\[2f(0)^2=f(2f(0))^2/2f(0)+f(0)\]](http://latex.artofproblemsolving.com/9/0/3/9035eba4b69d65cdbb8b1fa30ad8479576366d5d.png)
![\[2f(0)^2(2f(0)-1)=f(2f(0))^2\]](http://latex.artofproblemsolving.com/c/d/2/cd2917b8b090b60d048537a927504e3f29fc5cf6.png)
This means that 
is a perfect square. However, this is impossible, as it is equivalent to 
Therefore, 
Now substitute 
to get ![\[xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2.\]](http://latex.artofproblemsolving.com/a/8/6/a8601ea919c7e01aacfd97ce923e7128b2d7b869.png)
Similarly, ![\[x^2f(x)=f(-x)^2.\]](http://latex.artofproblemsolving.com/6/f/4/6f4ede18970bf6a7151c34e32a7de7fabaca23d2.png)
From these two equations, we can find either 
or 
Both of these are valid solutions on their own, so let's see if there are any solutions combining the two.
Let's say we can find 
and 
Then ![\[xf(-x)+y^2f(2x)=f(x)^2/x.\]](http://latex.artofproblemsolving.com/b/f/a/bfad0b72e3c79028d1487f996f6b987fc04cafb2.png)
![\[y^2f(2x)=x-x^3.\]](http://latex.artofproblemsolving.com/e/9/a/e9a809d6ac406460b6510efe3025392267ee9245.png)
(NEEDS FIXING: 
, so the RHS is 
instead of 
.)
If 
then 
which is only possible when 
This contradicts our assumption. Therefore, 
This forces 
due to the right side of the equation. Let's consider the possibility 
Substituting 
into the original equation yields ![\[0=2f(0)+1f(2)=0+f(1)=1,\]](http://latex.artofproblemsolving.com/0/2/4/024ce7bcceaf5c1e0f2c9d64c7e3e6a5c796fe8f.png)
which is impossible. So 
and there are no solutions "combining" 
and 
Therefore our only solutions are 
and 
