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

I will denote the original equation $f(x^2-y)+2yf(x)=f(f(x))+f(y)$ as OE.

I claim that the only solutions are $f(x) = -x^2, f(x) = 0,$ and $f(x) = x^2.$

Lemma 1: $f(0) = 0.$

Proof of Lemma 1:

We prove this by contradiction. Assume $f(0) = k \neq 0.$

By letting $x=y=0$ in the OE, we have $f(0) = f^2(0) + f(0) \Longrightarrow f^2(0) = 0 \Longrightarrow f(k) = 0.$

If we let $x = 0$ and $y = k^2$ in the OE, we have $f(-k^2) + 2k^2f(0) = f^2(0) + f(k^2) \Longrightarrow f(-k^2) + 2k^3 = f(k^2)$ and if we let $x = k$ and $y = k^2$ in the OE, we get

$f(0) + 2k^2f(k) = f^2(k) + f(k^2) \Longrightarrow k = k + f(k^2) \Longrightarrow f(k^2) = 0 \Longrightarrow f(-k^2) = 2k^3.$

However, upon substituting $x = k$ and $y = -k^2$ in the OE, this implies

$f(0) -2k^2f(k) = f^2(k) + f(-k^2) \Longrightarrow k = k + 2k^3 \Longrightarrow 2k^3 = 0.$

This means $k = 0,$ but we assumed $k \neq 0,$ contradiction, which proves the Lemma.

Substitute $y = 0$ in the OE to obtain $f(x^2) = f^2(x) + f(0) = f^2(x)$ and let $y = x^2$ in the OE to get $f(0) + 2x^2 f(x) = f^2(x) + f(x^2) = 2f(x^2) = 2x^2 f(x) \Longrightarrow \dfrac{f(x)}{x^2} = \dfrac{f(x^2)}{x^4} \Longrightarrow f(x) \propto x^2.$

Thus we can write $f(x) = kx^2$ for some $k.$ By $f(x^2) = f^2(x),$ we have $kx^4 = k^3x^4,$ so $k = -1, 0, 1,$ yielding the solutions $f(x) = -x^2, f(x) = 0, f(x) = x^2. \blacksquare$

- spectraldragon8

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

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Problem

Solution 1

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