# 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 15:30, 27 June 2019

## Problem

Let be the set of integers. Find all functions such that for all with .

## Solution

Let's assume Substitute to get

This means that is a perfect square. However, this is impossible, as it is equivalent to Therefore, Now substitute to get Similarly, 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 (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 which is impossible. So and there are no solutions "combining" and

Therefore our only solutions are and