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

(Created page with "==Problem== Let <math>\mathbb{R}</math> denote the set of all real numbers. Find all functions <math>f:\mathbb{R} \to \mathbb{R}</math> such that <cmath>f\left( x^{2}+f(y)...")
 
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==Solution==
 
==Solution==
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{{solution}}
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[quote=probability1.01]Set x = 0 to get <math>f(f(y)) = y+f(0)^{2}</math>. We'll let <math>c = f(0)^{2}</math>, so <math>f(f(y)) = y+c</math>. Then
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<math>f(a^{2}+f(f(b))) = f(b)+f(a)^{2}</math>
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<math>f(a^{2}+b+c) = f(b)+f(a)^{2}</math>
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<math>f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}</math>
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<math>a^{2}+b+2c = b+(a+c)^{2}</math>
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<math>2c = c^{2}+2ac</math>
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Since this holds for all <math>a</math>, it follows that <math>c = 0</math>. Now we have <math>f(0) = 0 \implies f(f(y)) = y</math>. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if <math>f(a) = f(b)</math>, then <math>a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}</math>. Finally, if <math>u > v</math>, then there is some <math>t</math> s.t. <math>u = t^{2}+v</math>, and so <math>f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)</math>. Hence f is strictly increasing. It is now clear that since <math>f(f(y)) = y</math>, we must have <math>f(x) = x</math> for all x.[/quote]
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==See Also==
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{{IMO box|year=1992|num-b=1|num-a=3}}
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[[Category:Olympiad Geometry Problems]]
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[[Category:3D Geometry Problems]]

Latest revision as of 13:53, 12 March 2024

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

[quote=probability1.01]Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then

$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$ $f(a^{2}+b+c) = f(b)+f(a)^{2}$ $f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$ $a^{2}+b+2c = b+(a+c)^{2}$ $2c = c^{2}+2ac$

Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.[/quote]

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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