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

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(→‎Solution 1)
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-codemaster11
 
-codemaster11
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== Solution 2 ==
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Let our equation be <math>P(x,y)</math>. We start by plugging in some initial values:
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<math>y=x^2:\; f(0)+2x^2f(x) = f(f(x))+f(x^2) \;\;\;\; (1)</math>
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<math>y=0:\; f(x^2) = f(f(x))+f(0) \;\;\;\; (2)</math>
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<math>x=0:\; f(-y) + 2yf(0) = f(f(0)) + f(y) \;\;\;\; (3)</math>
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Plugging in <math>x=1</math> into <math>(3)</math> gives
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<cmath>f(1) = f(f(1)) + f(0) \;\;\;\; (4).</cmath>
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From <math>(1)</math>, we get
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<cmath>f(0) + 2x^2f(x) = 2f(x^2)-f(0) \implies x^2f(x)+f(0) = f(x^2)</cmath>
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Substituting in what we have in <math>(3)</math> gives
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<cmath>x^2f(x)+f(0) = f(0) = f(f(x)) \implies x^2f(x) = f(f(x)).</cmath>
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Plugging in <math>x=1</math> gives
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<cmath>f(1)=f(f(1)) \;\;\;\; (5).</cmath>
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As a result, <math>(4)</math> becomes <math>f(0)=0</math>.
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Now, <math>(3)</math> becomes
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<cmath>f(x^2) = f(f(x)) \;\;\;\; (6)</cmath>
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and <math>(2)</math> becomes
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<cmath>f(y)=f(-y) \;\;\;\; (7).</cmath>
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Note that <math>f\equiv 0</math> is a solution. Now, assume <math>f(x) \neq 0</math>.
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<b>Claim:</b> <math>f</math> is injective over <math>\mathbb{R}^{+}</math>.
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Let <math>f(a) = f(b) \neq 0</math> with <math>a,b>0</math>. Plugging in <math>x=a, y=b^2</math> and <math>x=b, y=a^2</math> into <math>P</math> gives us
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<cmath>f(a^2-b^2)+2b^2f(a) = f(a^2)+f(b^2)</cmath>
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<cmath>f(b^2-a^2)+2a^2f(b) = f(b^2)+f(a^2)</cmath>
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Subtracting, and using <math>(7)</math> gives us <math>2(a^2-b^2)f(a) = 0</math>, which implies that either <math>f(a)=0</math> or <math>a=\pm b</math>. Either way leads to contradiction. Thus, <math>f</math> is injective. <math>\square</math>
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As a result, <math>(6)</math> becomes <math>f(x)=\pm x^2</math>. Piecing everything yields <math>f(x) = 0, \pm x^2</math>.
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It just remains to verify these solutions work, and doing so is quite trivial;
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<cmath>f(x)=0:\; 0+0 = 0+0,</cmath>
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<cmath>f(x)=x^2:\; (x^2-y)^2+2yx^2 = x^4+y^2,</cmath>
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<cmath>f(x)=-x^2:\; -(x^2-y)^2-2yx^2 = -x^4-y^2,</cmath>
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all of which are obviously true.
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~sml1809
  
 
==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 13:53, 28 June 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

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

Solution 2

Let our equation be $P(x,y)$. We start by plugging in some initial values:

$y=x^2:\; f(0)+2x^2f(x) = f(f(x))+f(x^2) \;\;\;\; (1)$

$y=0:\; f(x^2) = f(f(x))+f(0) \;\;\;\; (2)$

$x=0:\; f(-y) + 2yf(0) = f(f(0)) + f(y) \;\;\;\; (3)$

Plugging in $x=1$ into $(3)$ gives \[f(1) = f(f(1)) + f(0) \;\;\;\; (4).\] From $(1)$, we get \[f(0) + 2x^2f(x) = 2f(x^2)-f(0) \implies x^2f(x)+f(0) = f(x^2)\] Substituting in what we have in $(3)$ gives \[x^2f(x)+f(0) = f(0) = f(f(x)) \implies x^2f(x) = f(f(x)).\] Plugging in $x=1$ gives \[f(1)=f(f(1)) \;\;\;\; (5).\] As a result, $(4)$ becomes $f(0)=0$.

Now, $(3)$ becomes \[f(x^2) = f(f(x)) \;\;\;\; (6)\] and $(2)$ becomes \[f(y)=f(-y) \;\;\;\; (7).\] Note that $f\equiv 0$ is a solution. Now, assume $f(x) \neq 0$.

Claim: $f$ is injective over $\mathbb{R}^{+}$.

Let $f(a) = f(b) \neq 0$ with $a,b>0$. Plugging in $x=a, y=b^2$ and $x=b, y=a^2$ into $P$ gives us \[f(a^2-b^2)+2b^2f(a) = f(a^2)+f(b^2)\] \[f(b^2-a^2)+2a^2f(b) = f(b^2)+f(a^2)\] Subtracting, and using $(7)$ gives us $2(a^2-b^2)f(a) = 0$, which implies that either $f(a)=0$ or $a=\pm b$. Either way leads to contradiction. Thus, $f$ is injective. $\square$

As a result, $(6)$ becomes $f(x)=\pm x^2$. Piecing everything yields $f(x) = 0, \pm x^2$.

It just remains to verify these solutions work, and doing so is quite trivial; \[f(x)=0:\; 0+0 = 0+0,\] \[f(x)=x^2:\; (x^2-y)^2+2yx^2 = x^4+y^2,\] \[f(x)=-x^2:\; -(x^2-y)^2-2yx^2 = -x^4-y^2,\] all of which are obviously true.

~sml1809

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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