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

(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))=\...")
 
(Added solution)
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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==
 
==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: <math>f(0) = 0</math>.
 +
Proof: Assume the opposite for a contradiction. Plug in <math>x = 2f(0)</math> (because we assumed that <math>f(0) \neq 0</math>), <math>y = 0</math>. What you get eventually reduces to:
 +
<cmath>4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2</cmath>
 +
which is a contradiction since the RHS is divisible by 2 but not 4.
 +
 +
Then plug in <math>y = 0</math> into the original equation and simplify by Lemma 1. We get:
 +
<cmath>x^2f(-x) = f(x)^2</cmath>
 +
Then:
 +
\[
 +
\begin{align*}
 +
x^6f(x) &= x^4(-x)^2f(-(-x)) \\
 +
&= x^4f(-x)^2 \\
 +
&= f(x)^4
 +
\end{align*}
 +
\]
 +
Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>.
 +
 +
Now either <math>f(x)</math> is <math>x^2</math> for all <math>x</math> or there exists <math>a \neq 0</math> such that <math>f(a)=0</math>. The first case gives a valid solution. In the second case, we let <math>y = a</math> in the original equation and simplify to get:
 +
<cmath>xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}</cmath>
 +
But we know that <math>xf(-x) = \frac{f(x)^2}{x}</math>, so:
 +
<cmath>a^2f(2x) = 0</cmath>
 +
Since <math>a</math> is not 0, <math>f(2x)</math> is 0 for all <math>x</math> (including 0). Now either <math>f(x)</math> is 0 for all <math>x</math>, or there exists some <math>m \neq 0</math> such that <math>f(m) = m^2</math>. Then <math>m</math> must be odd. We can let <math>x = 2k</math> in the original equation, and since <math>f(2x)</math> is 0 for all <math>x</math>, stuff cancels and we get:
 +
<cmath>y^2f(4k - f(y)) = f(yf(y))</cmath>
 +
[b]for <math>k \neq 0</math>.[/b]
 +
Now, let <math>y = m</math> and we get:
 +
<cmath>m^2f(4k - m^2) = f(m^3)</cmath>
 +
Now, either both sides are 0 or both are equal to <math>m^6</math>. If both are <math>m^6</math> then:
 +
<cmath>m^2(4k - m^2)^2 = m^6</cmath>
 +
which simplifies to:
 +
<cmath>4k - m^2 = \pm m^2</cmath>
 +
Since <math>k \neq 0</math> and <math>m</math> is odd, both cases are impossible, so we must have:
 +
<cmath>m^2f(4k - m^2) = f(m^3) = 0</cmath>
 +
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, so our only solutions are <math>f(x) = 0</math> and <math>f(x) = x^2</math>.

Revision as of 18:40, 29 April 2014

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

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: $f(0) = 0$. Proof: Assume the opposite for a contradiction. Plug in $x = 2f(0)$ (because we assumed that $f(0) \neq 0$), $y = 0$. What you get eventually reduces to: \[4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2\] which is a contradiction since the RHS is divisible by 2 but not 4.

Then plug in $y = 0$ into the original equation and simplify by Lemma 1. We get: \[x^2f(-x) = f(x)^2\] Then: \[ \begin{align*} x^6f(x) &= x^4(-x)^2f(-(-x)) \\ &= x^4f(-x)^2 \\ &= f(x)^4 \end{align*} \] Therefore, $f(x)$ must be 0 or $x^2$.

Now either $f(x)$ is $x^2$ for all $x$ or there exists $a \neq 0$ such that $f(a)=0$. The first case gives a valid solution. In the second case, we let $y = a$ in the original equation and simplify to get: \[xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}\] But we know that $xf(-x) = \frac{f(x)^2}{x}$, so: \[a^2f(2x) = 0\] Since $a$ is not 0, $f(2x)$ is 0 for all $x$ (including 0). Now either $f(x)$ is 0 for all $x$, or there exists some $m \neq 0$ such that $f(m) = m^2$. Then $m$ must be odd. We can let $x = 2k$ in the original equation, and since $f(2x)$ is 0 for all $x$, stuff cancels and we get: \[y^2f(4k - f(y)) = f(yf(y))\] [b]for $k \neq 0$.[/b] Now, let $y = m$ and we get: \[m^2f(4k - m^2) = f(m^3)\] Now, either both sides are 0 or both are equal to $m^6$. If both are $m^6$ then: \[m^2(4k - m^2)^2 = m^6\] which simplifies to: \[4k - m^2 = \pm m^2\] Since $k \neq 0$ and $m$ is odd, both cases are impossible, so we must have: \[m^2f(4k - m^2) = f(m^3) = 0\] Then we can let $k$ be anything except 0, and get $f(x)$ is 0 for all $x \equiv 3 \pmod{4}$ except $-m^2$. Also since $x^2f(-x) = f(x)^2$, we have $f(x) = 0 \Rightarrow f(-x) = 0$, so $f(x)$ is 0 for all $x \equiv 1 \pmod{4}$ except $m^2$. So $f(x)$ is 0 for all $x$ except $\pm m^2$. Since $f(m) \neq 0$, $m = \pm m^2$. Squaring, $m^2 = m^4$ and dividing by $m$, $m = m^3$. Since $f(m^3) = 0$, $f(m) = 0$, which is a contradiction, so our only solutions are $f(x) = 0$ and $f(x) = x^2$.

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