# Difference between revisions of "2014 USAMO Problems/Problem 2"

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Then we can let <math>k</math> be anything except 0, and get <math>f(x)</math> is 0 for all <math>x \equiv 3 \pmod{4}</math> except <math>-m^2</math>. Also since <math>x^2f(-x) = f(x)^2</math>, we have <math>f(x) = 0 \Rightarrow f(-x) = 0</math>, so <math>f(x)</math> is 0 for all <math>x \equiv 1 \pmod{4}</math> except <math>m^2</math>. So <math>f(x)</math> is 0 for all <math>x</math> except <math>\pm m^2</math>. Since <math>f(m) \neq 0</math>, <math>m = \pm m^2</math>. Squaring, <math>m^2 = m^4</math> and dividing by <math>m</math>, <math>m = m^3</math>. Since <math>f(m^3) = 0</math>, <math>f(m) = 0</math>, which is a contradiction for <math>m \neq 1</math>. However, if we plug in <math>x = 1</math> with <math>f(1) = 1</math> and <math>y</math> as an arbitrary large number with <math>f(y) = 0</math> into the original equation, we get <math>0 = 1</math> which is a clear contradiction, so our only solutions are <math>f(x) = 0</math> and <math>f(x) = x^2</math>. | Then we can let <math>k</math> be anything except 0, and get <math>f(x)</math> is 0 for all <math>x \equiv 3 \pmod{4}</math> except <math>-m^2</math>. Also since <math>x^2f(-x) = f(x)^2</math>, we have <math>f(x) = 0 \Rightarrow f(-x) = 0</math>, so <math>f(x)</math> is 0 for all <math>x \equiv 1 \pmod{4}</math> except <math>m^2</math>. So <math>f(x)</math> is 0 for all <math>x</math> except <math>\pm m^2</math>. Since <math>f(m) \neq 0</math>, <math>m = \pm m^2</math>. Squaring, <math>m^2 = m^4</math> and dividing by <math>m</math>, <math>m = m^3</math>. Since <math>f(m^3) = 0</math>, <math>f(m) = 0</math>, which is a contradiction for <math>m \neq 1</math>. However, if we plug in <math>x = 1</math> with <math>f(1) = 1</math> and <math>y</math> as an arbitrary large number with <math>f(y) = 0</math> into the original equation, we get <math>0 = 1</math> which is a clear contradiction, so our only solutions are <math>f(x) = 0</math> and <math>f(x) = x^2</math>. | ||

+ | ==Alternative Solution== | ||

+ | Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and <math>f(yf(y))</math> must also be an integer, therefore <math>\frac{f(x)^2}{x}</math> is an integer. If <math>x</math> divides <math>f(x)^2</math> for all integers <math>x \ne 0</math>, then <math>x</math> must be a factor of <math>f(x)</math>, therefore <math>f(0)=0</math>. Now, by setting <math>y=0</math> in the original equation, we have <math>xf(-x)=\frac{f(x)^2}{x}</math>. Assuming <math>x \ne 0</math>, we have <math>x^2f(-x)=f(x)^2</math>. Substituting in <math>-x</math> for <math>x</math> gives us <math>x^2f(x)=f(-x)^2</math>. Substituting in <math>\frac{f(x)^2}{x^2}</math> in for <math>f(-x)</math> in the second equation gives us <math>x^2f(x)=\frac{f(x)^4}{x^4}</math>, so <math>x^6f(x)=f(x)^4</math>. In particular, if <math>f(x) \ne 0</math>, then we have <math>f(x)^3=x^6</math>, therefore <math>f(x)=0, x^2</math> are the only possible solutions. | ||

[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||

[[Category:Functional Equation Problems]] | [[Category:Functional Equation Problems]] |

## Revision as of 11:13, 4 January 2019

## 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 LHS is divisible by 2 but not 4.

Then plug in into the original equation and simplify by Lemma 1. We get: Then:

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 for . However, if we plug in with and as an arbitrary large number with into the original equation, we get which is a clear contradiction, so our only solutions are and .

## Alternative Solution

Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and must also be an integer, therefore is an integer. If divides for all integers , then must be a factor of , therefore . Now, by setting in the original equation, we have . Assuming , we have . Substituting in for gives us . Substituting in in for in the second equation gives us , so . In particular, if , then we have , therefore are the only possible solutions.