Difference between revisions of "2014 USAJMO Problems/Problem 3"

(Created page with "==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))=\...")
 
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==Solution==
 
==Solution==
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Let's assume <math>f(0)\neq 0.</math> Substitute <math>(x,y)=(2f(0),0)</math> to get <cmath>2f(0)^2=f(2f(0))^2/2f(0)+f(0)</cmath>
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<cmath>2f(0)^2(2f(0)-1)=f(2f(0))^2</cmath>
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This means that <math>2(2f(0)-1)</math> is a perfect square. However, this is impossible, as it is equivalent to <math>2\pmod{4}.</math> Therefore, <math>f(0)=0.</math> Now substitute <math>x\neq 0, y=0</math> to get <cmath>xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2.</cmath>
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Similarly, <cmath>x^2f(x)=f(-x)^2.</cmath>
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From these two equations, we can find either <math>f(x)=f(-x)=0,</math> or <math>f(x)=f(-x)=x^2.</math> Both of these are valid solutions on their own, so let's see if there are any solutions combining the two.
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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>
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<cmath>y^2f(2x)=x-x^3.</cmath>
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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>
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Therefore our only solutions are <math>\boxed{f(x)=0}</math> and <math>\boxed{f(x)=x^2.}</math>

Revision as of 20:04, 18 April 2016

Problem

Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that \[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\] for all $x, y \in \mathbb{Z}$ with $x \neq 0$.

Solution

Let's assume $f(0)\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get \[2f(0)^2=f(2f(0))^2/2f(0)+f(0)\] \[2f(0)^2(2f(0)-1)=f(2f(0))^2\]

This means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\neq 0, y=0$ to get \[xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2.\] Similarly, \[x^2f(x)=f(-x)^2.\] From these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ 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 $f(x)=x^2, f(y)=0,$ and $x,y\neq 0.$ Then \[xf(-x)+y^2f(2x)=f(x)^2/x.\] \[y^2f(2x)=x-x^3.\]

If $f(2x)=4x^2,$ then $y^2=\frac{x-x^3}{4x^2}=\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields \[0=2f(0)+1f(2)=0+f(1)=1,\] which is impossible. So $f(2)=f(-2)=4$ and there are no solutions "combining" $f(x)=x^2$ and $f(x)=0.$

Therefore our only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x^2.}$

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