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: $y^2f(4k - f(y)) = f(yf(y))$ for $\mathbf{k \neq 0}$ . 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.

Solution 3

Let's assume $f(0)\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get $2f(0)^2=f(2f(0))^2/2f(0)+f(0)$ $2f(0)^2(2f(0)-1)=f(2f(0))^2$

This means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\neq 0, y=0$ to get $xf(-x)=\frac{f(x)^2}{x} \implies x^2f(-x)=f(x)^2.$ Similarly, $x^2f(x)=f(-x)^2.$ From these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ 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 $f(x)=x^2, f(y)=0,$ and $x,y\neq 0.$ Then $xf(-x)+y^2f(2x)=f(x)^2/x.$ $y^2f(2x)=x-x^3.$ (NEEDS FIXING: $f(x)^2/x= x^4/x = x^3$ , so the RHS is $0$ instead of $x-x^3$ .)

If $f(2x)=4x^2,$ then $y^2=\frac{x-x^3}{4x^2}=\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields $0=2f(0)+1f(2)=0+f(1)=1,$ which is impossible. So $f(2)=f(-2)=4$ and there are no solutions "combining" $f(x)=x^2$ and $f(x)=0.$

Therefore our only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x^2.}$

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2014 USAMO Problems/Problem 2

Contents

Problem

Solution

Alternative Solution

Solution 3