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