Difference between revisions of "2014 USAMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
Let <math>\mathbb{Z}</math> be the set of integers. Find all functions <math>f : \mathbb{Z} \rightarrow \mathbb{Z}</math> such that <cmath>xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))</cmath> for all <math>x, y \in \mathbb{Z}</math> with <math>x \neq 0</math>. | Let <math>\mathbb{Z}</math> be the set of integers. Find all functions <math>f : \mathbb{Z} \rightarrow \mathbb{Z}</math> such that <cmath>xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))</cmath> for all <math>x, y \in \mathbb{Z}</math> with <math>x \neq 0</math>. | ||
− | ==Solution== | + | <math>Insert formula here</math>==Solution== |
Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions. | Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions. | ||
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<cmath>x^2f(-x) = f(x)^2</cmath> | <cmath>x^2f(-x) = f(x)^2</cmath> | ||
Then: | Then: | ||
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− | \begin{align*} | + | <math>\begin{align*} |
x^6f(x) &= x^4(-x)^2f(-(-x)) \\ | x^6f(x) &= x^4(-x)^2f(-(-x)) \\ | ||
&= x^4f(-x)^2 \\ | &= x^4f(-x)^2 \\ | ||
&= f(x)^4 | &= f(x)^4 | ||
− | \end{align*} | + | \end{align*}</math> |
− | + | ||
Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>. | Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>. | ||
Revision as of 17:40, 29 April 2014
Problem
Let be the set of integers. Find all functions such that for all with . ==Solution== Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.
Lemma 1: . Proof: Assume the opposite for a contradiction. Plug in (because we assumed that ), . What you get eventually reduces to: which is a contradiction since the RHS is divisible by 2 but not 4.
Then plug in into the original equation and simplify by Lemma 1. We get: Then:
$\begin{align*} x^6f(x) &= x^4(-x)^2f(-(-x)) \\ &= x^4f(-x)^2 \\ &= f(x)^4 \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore, must be 0 or .
Now either is for all or there exists such that . The first case gives a valid solution. In the second case, we let in the original equation and simplify to get: But we know that , so: Since is not 0, is 0 for all (including 0). Now either is 0 for all , or there exists some such that . Then must be odd. We can let in the original equation, and since is 0 for all , stuff cancels and we get: [b]for .[/b] Now, let and we get: Now, either both sides are 0 or both are equal to . If both are then: which simplifies to: Since and is odd, both cases are impossible, so we must have: Then we can let be anything except 0, and get is 0 for all except . Also since , we have , so is 0 for all except . So is 0 for all except . Since , . Squaring, and dividing by , . Since , , which is a contradiction, so our only solutions are and .