2024 USAJMO Problems/Problem 5: Difference between revisions

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

@@ Line 8: / Line 8: @@
 == Solution 1 ==
+I will denote the original equation <math>f(x^2-y)+2yf(x)=f(f(x))+f(y)</math> as OE.
+I claim that the only solutions are <math>f(x) = -x^2, f(x) = 0,</math> and <math>f(x) = x^2.</math>
+Lemma 1: <math>f(0) = 0.</math>
+Proof of Lemma 1:
+We prove this by contradiction. Assume <math>f(0) = k \neq 0.</math>
+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>
+If we let <math>x = 0</math> and <math>y = k^2</math> in the OE, we have
+<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
+<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>
+However, upon substituting <math>x = k</math> and <math>y = -k^2</math> in the OE, this implies
+<cmath>f(0) -2k^2f(k) = f^2(k) + f(-k^2) \Longrightarrow k = k + 2k^3 \Longrightarrow 2k^3 = 0.</cmath>
+This means <math>k = 0,</math> but we assumed <math>k \neq 0,</math> contradiction, which proves the Lemma.
+Substitute <math>y = 0</math> in the OE to obtain
+<cmath>f(x^2) = f^2(x) + f(0) = f^2(x)</cmath>
+and let <math>y = x^2</math> in the OE to get
+<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>
+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>
+- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]
 ==See Also==
 {{USAJMO newbox|year=2024|num-b=4|num-a=6}}
 {{MAA Notice}}

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2024 USAJMO Problems/Problem 5: Difference between revisions

Revision as of 12:04, 24 March 2024

Contents

Problem

Solution 1

See Also