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

Revision as of 19:49, 24 March 2015

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 LHS 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*} $x^6f(x) &= x^4(-x)^2f(-(-x)) \\ &= x^4f(-x)^2 \\ &= f(x)^4$ (Error compiling LaTeX. Unknown error_msg) \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 $f(x) = x^2$ .

@@ Line 7: / Line 7: @@
 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.
+which is a contradiction since the LHS 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:
@@ Line 13: / Line 13: @@
 Then:
-<math>\begin{align*}
+\begin{align*}
-x^6f(x) &= x^4(-x)^2f(-(-x)) \\
+<math>x^6f(x) &= x^4(-x)^2f(-(-x)) \\
 &= x^4f(-x)^2 \\
-&= f(x)^4
+&= f(x)^4</math>
-\end{align*}</math>
+\end{align*}
 Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>.

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

Revision as of 19:49, 24 March 2015

Problem

Solution