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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]
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| | ==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}} |
that satisfy
![\[f(x^2-y)+2yf(x)=f(f(x))+f(y)\]](http://latex.artofproblemsolving.com/4/b/d/4bdeaa49c29c972b00378da74fa56558da3d543e.png)
for all 
.
