Difference between revisions of "1992 IMO Problems/Problem 2"

Revision as of 10:23, 20 November 2023

Problem

Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that

$f\left( x^{2}+f(y) \right)= y+(f(x))^{2} \hspace{0.5cm} \forall x,y \in \mathbb{R}$

Solution

We notice that the right hand side of the equation has $y$ , therefore the only way that $f\left( x^{2}+f(y) \right)$ produces that $y$ is if $f(y)=y$ .

This makes the equation as $f\left( x^{2}+y \right)= y+(f(x))^{2}$

Since $f(y)=y$ , then $f(x)=x$ , thus $f\left( x^{2}+y \right)=x^{2}+y=y+(f(x))^{2}=y+x^{2}$ and the equation holds true.

Only solution to this problem is $f(x)=x$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 6: / Line 6: @@
 ==Solution==
-{{solution}}
+We notice that the right hand side of the equation has <math>y</math>, therefore the only way that <math>f\left( x^{2}+f(y) \right)</math> produces that <math>y</math> is if <math>f(y)=y</math>.
+This makes the equation as <math>f\left( x^{2}+y \right)= y+(f(x))^{2}</math>
+Since <math>f(y)=y</math>, then <math>f(x)=x</math>, thus <math>f\left( x^{2}+y \right)=x^{2}+y=y+(f(x))^{2}=y+x^{2}</math> and the equation holds true.
+Only solution to this problem is <math>f(x)=x</math>
+{{alternate solutions}}
 ==See Also==

1992 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions

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Difference between revisions of "1992 IMO Problems/Problem 2"

Revision as of 10:23, 20 November 2023

Problem

Solution

See Also