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Difference between revisions of "2024 USAJMO Problems/Problem 5"

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== Solution 1 ==
 
== Solution 1 ==
  
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I will denote the original equation <math>f(x^2-y)+2yf(x)=f(f(x))+f(y)</math> as OE.
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I claim that the only solutions are <math>f(x) = -x^2, f(x) = 0,</math> and <math>f(x) = x^2.</math>
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Lemma 1: <math>f(0) = 0.</math>
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Proof of Lemma 1:
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We prove this by contradiction. Assume <math>f(0) = k \neq 0.</math>
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By letting <math>x=y=0</math> in the OE, we have <cmath>f(0) = f^2(0) + f(0) \Longrightarrow f^2(0) = 0 \Longrightarrow f(k) = 0.</cmath>
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If we let <math>x = 0</math> and <math>y = k^2</math> in the OE, we have
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<cmath>f(-k^2) + 2k^2f(0) = f^2(0) + f(k^2) \Longrightarrow f(-k^2) + 2k^3 = f(k^2)</cmath> and if we let <math>x = k</math> and <math>y = k^2</math> in the OE, we get
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<cmath>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. </cmath>
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However, upon substituting <math>x = k</math> and <math>y = -k^2</math> in the OE, this implies
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<cmath>f(0) -2k^2f(k) = f^2(k) + f(-k^2) \Longrightarrow k = k + 2k^3 \Longrightarrow 2k^3 = 0.</cmath>
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This means <math>k = 0,</math> but we assumed <math>k \neq 0,</math> contradiction, which proves the Lemma.
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Substitute <math>y = 0</math> in the OE to obtain
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<cmath>f(x^2) = f^2(x) + f(0) = f^2(x)</cmath>
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and let <math>y = x^2</math> in the OE to get
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<cmath>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.</cmath>
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Thus we can write <math>f(x) = kx^2</math> for some <math>k.</math> By <math>f(x^2) = f^2(x),</math> we have <cmath>kx^4 = k^3x^4,</cmath> so <math>k = -1, 0, 1,</math> yielding the solutions <cmath>f(x) = -x^2, f(x) = 0, f(x) = x^2.  \blacksquare</cmath>
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- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]
  
 
==See Also==
 
==See Also==
 
{{USAJMO newbox|year=2024|num-b=4|num-a=6}}
 
{{USAJMO newbox|year=2024|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:04, 24 March 2024

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

See Also

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

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