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

(→‎Solution 1)
(→‎Solution 1)
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Plugging in <math>y</math> as <math>0:</math>
 
Plugging in <math>y</math> as <math>0:</math>
 
\begin{equation}
 
\begin{equation}
−
f(x^2)=f(f(x))+f(0)  
+
f(x^2)=f(f(x))+f(0) \text{ } (1)
 
\end{equation}
 
\end{equation}
 
Plugging in <math>x, y</math> as <math>0:</math>
 
Plugging in <math>x, y</math> as <math>0:</math>
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but since <math>f(f(0))=0,</math>
 
but since <math>f(f(0))=0,</math>
 
\begin{equation}
 
\begin{equation}
−
f(-y)+2yf(0)=f(y)
+
f(-y)+2yf(0)=f(y) \text{ } (2)
 
\end{equation}
 
\end{equation}
 
Plugging in <math>y^2</math> instead of <math>y</math> in the given equation:
 
Plugging in <math>y^2</math> instead of <math>y</math> in the given equation:
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The difference would be:
 
The difference would be:
 
\begin{equation}
 
\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)
+
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}
 
\end{equation}
 
The right-hand side would be <math>f(0)-f(0)=0</math> by <math>(1).</math> Also,
 
The right-hand side would be <math>f(0)-f(0)=0</math> by <math>(1).</math> Also,

Revision as of 21:57, 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) \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

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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