Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 13: Line 13:
 
Then:
 
Then:
  
−
<cmath>x^6f(x) = x^4(-x)^2f(-(-x))</cmath>
+
<cmath>\begin{align*}
−
<cmath>= x^4f(-x)^2</cmath>
+
x^6f(x) = x^4(-x)^2f(-(-x))\\
−
<cmath>= f(x)^4</cmath>
+
&= x^4f(-x)^2\\
 +
&= f(x)^4
 +
\end{align*}</cmath>
  
 
Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>.
 
Therefore, <math>f(x)</math> must be 0 or <math>x^2</math>.

Revision as of 19:57, 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 \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 $f(x) = x^2$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&oldid=69465"