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

(→‎Solution 1)
(→‎Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
 +
Plugging in <math>y</math> as <math>0:</math>
 +
\begin{equation}
 +
f(x^2)=f(f(x))+f(0)
 +
\end{equation}
 +
Plugging in <math>x, y</math> as <math>0:</math>
 +
<cmath>f(0)=f(f(0))+f(0)</cmath>
 +
or
 +
<cmath>f(f(0))=0</cmath>
 +
Plugging in <math>x</math> as <math>0:</math>
 +
<cmath>f(-y)+2yf(0)=f(f(0))+f(y),</cmath>
 +
but since <math>f(f(0))=0,</math>
 +
\begin{equation}
 +
f(-y)+2yf(0)=f(y)
 +
\end{equation}
 +
Plugging in <math>y^2</math> instead of <math>y</math> in the given equation:
 +
<cmath>f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)</cmath>
 +
Replacing <math>y</math> and <math>x</math>:
 +
<cmath>f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)</cmath>
 +
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)
 +
\end{equation}
 +
The right-hand side would be <math>f(0)-f(0)=0</math> by <math>(1).</math> Also,
 +
<cmath>f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)</cmath> by <math>(2)</math>
 +
So, <math>(3)</math> is reduced to:
 +
<cmath>2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0</cmath>
 +
Regrouping and dividing by 2:
 +
<cmath>y^2(f(x)-f(0))=x^2(f(y)-f(0))</cmath>
 +
<cmath>\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}</cmath>
 +
Because this holds for all x and y, <math>\frac{f(x)-f(0)}{x^2}</math> is a constant. So, <math>f(x)=cx^2+f(0)</math>.
 +
This function must be even, so <math>f(y)-f(-y)=0</math>.
 +
So, along with <math>(2)</math>, <math>2yf(0)=0</math> for all <math>y</math>, so <math>f(0)=0</math>, and <math>f(x)=cx^2</math>.
 +
Plugging in <math>cx^2</math> for <math>f(x)</math> in the original equation, we get:
 +
<cmath>c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2</cmath>
 +
<cmath>c(x^4+y^2)=c(c^2x^4+y^2)</cmath>
 +
So, <math>c=0</math> or <math>c^2=1.</math>
 +
All of these solutions work, so the solutions are <math>f(x)=-x^2, 0, x^2</math>.
 +
 +
-codemaster11
  
 
==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 21:55, 5 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) \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) \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) \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

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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