2024 USAJMO Problems/Problem 5

Problem

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy $f(x^2-y)+2yf(x)=f(f(x))+f(y)$ for all $x,y\in\mathbb{R}$ .

Solution 1

Plugging in $y$ as $0:$ \begin{equation} f(x^2)=f(f(x))+f(0) \text{ } (1) \end{equation} Plugging in $x, y$ as $0:$ $f(0)=f(f(0))+f(0)$ or $f(f(0))=0$ Plugging in $x$ as $0:$ $f(-y)+2yf(0)=f(f(0))+f(y),$ but since $f(f(0))=0,$ \begin{equation} f(-y)+2yf(0)=f(y) \text{ } (2) \end{equation} Plugging in $y^2$ instead of $y$ in the given equation: $f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$ Replacing $y$ and $x$ : $f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)$ The difference would be: \begin{equation} f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } (3) \end{equation} The right-hand side would be $f(0)-f(0)=0$ by $(1).$ Also, $f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)$ by $(2)$ So, $(3)$ is reduced to: $2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0$ Regrouping and dividing by 2: $y^2(f(x)-f(0))=x^2(f(y)-f(0))$ $\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}$ Because this holds for all x and y, $\frac{f(x)-f(0)}{x^2}$ is a constant. So, $f(x)=cx^2+f(0)$ . This function must be even, so $f(y)-f(-y)=0$ . So, along with $(2)$ , $2yf(0)=0$ for all $y$ , so $f(0)=0$ , and $f(x)=cx^2$ . Plugging in $cx^2$ for $f(x)$ in the original equation, we get: $c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2$ $c(x^4+y^2)=c(c^2x^4+y^2)$ So, $c=0$ or $c^2=1.$ All of these solutions work, so the solutions are $f(x)=-x^2, 0, x^2$ .

-codemaster11

2024 USAJMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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2024 USAJMO Problems/Problem 5

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Problem

Solution 1

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