# 2014 USAMO Problems/Problem 2

## 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 for $m \neq 1$. However, if we plug in $x = 1$ with $f(1) = 1$ and $y$ as an arbitrary large number with $f(y) = 0$ into the original equation, we get $0 = 1$ which is a clear contradiction, so our only solutions are $f(x) = 0$ and $f(x) = x^2$.

## Alternative Solution

Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and $f(yf(y))$ must also be an integer, therefore $\frac{f(x)^2}{x}$ is an integer. If $x$ divides $f(x)^2$ for all integers $x \ne 0$, then $x$ must be a factor of $f(x)$, therefore $f(0)=0$. Now, by setting $y=0$ in the original equation, this simplifies to $xf(-x)=\frac{f(x)^2}{x}$. Assuming $x \ne 0$, we have $x^2f(-x)=f(x)^2$. Substituting in $-x$ for $x$ gives us $x^2f(x)=f(-x)^2$. Substituting in $\frac{f(x)^2}{x^2}$ in for $f(-x)$ in the second equation gives us $x^2f(x)=\frac{f(x)^4}{x^4}$, so $x^6f(x)=f(x)^4$. In particular, if $f(x) \ne 0$, then we have $f(x)^3=x^6$, therefore $f(x)$ is equivalent to $0$ or $x^2$ for every integer $x$. Now, we shall prove that if for some integer $t \ne 0$, if $f(t)=0$, then $f(x)=0$ for all integers $x$. If we assume $f(y)=0$ and $y \ne 0$ in the original equation, this simplifies to $xf(-x)+y^2f(2x)=\frac{f(x)^2}{x}$. However, since $x^2f(-x)=f(x)^2$, we can rewrite this equation as $\frac{f(x)^2}{x}+y^2f(2x)=\frac{f(x)^2}{x}$, $y^2f(2x)$ must therefore be equivalent to $0$. Since, by our initial assumption, $y \ne 0$, this means that $f(2x)=0$, so, if for some integer $y \ne 0$, $f(y)=0$, then $f(x)=0$ for all integers $x$. The contrapositive must also be true, i.e. If $f(x) \ne 0$ for all integers $x$, then there is no integral value of $y \ne 0$ such that $f(y)=0$, therefore $f(x)$ must be equivalent for $x^2$ for every integer $x$, including $0$, since $f(0)=0$. Thus, $f(x)=0, x^2$ are the only possible solutions.