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Difference between revisions of "2015 IMO Problems/Problem 5"

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==Problem==
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Let <math>\mathbb{R}</math> be the set of real numbers. Determine all functions <math>f</math>:<math>\mathbb{R}\rightarrow\mathbb{R}</math> satisfying the equation
 
Let <math>\mathbb{R}</math> be the set of real numbers. Determine all functions <math>f</math>:<math>\mathbb{R}\rightarrow\mathbb{R}</math> satisfying the equation
  
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Proposed by Dorlir Ahmeti, Albania
 
Proposed by Dorlir Ahmeti, Albania
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==Solution==
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<math>f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)</math> for all real numbers <math>x</math> and <math>y</math>.
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(1) Put <math>x=y=0</math> in the equation,
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We get<math> f(0 + f(0)) + f(0) = 0 + f(0) + 0</math>
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or <math>f(f(0)) = 0</math>
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Let <math>f(0) = k</math>, then <math>f(k) = 0</math>
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(2) Put <math>x=0, y=k</math> in the equation,
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We get <math>f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)</math>
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But <math>f(k) = 0</math> and <math>f(0) = k</math>
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so, <math>f(0) + f(0) = f(0)^2</math>
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or <math>f(0)[f(0) - 2] = 0</math>
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Hence <math>f(0) = 0, 2</math>
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Case <math>1</math> : <math>f(0) = 0</math>
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Put <math>x=0, y=x</math> in the equation,
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We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
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or, <math>f(f(x)) = f(x)</math>
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Say <math>f(x) = z</math>, we get <math>f(z) = z</math>
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So, <math>f(x) = x</math> is a solution -- fallacy
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Case <math>2</math> : <math>f(0) = 2</math>
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Again put <math>x=0, y=x</math> in the equation,
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We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
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or, <math>f(f(x)) + 2 = f(x) + 2x</math>
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We observe that <math>f(x)</math> must be a polynomial of power <math>1</math> as any other power (for that matter, any other function) will make the <math>LHS</math> and <math>RHS</math> of different powers and will not have any non-trivial solutions. -- fallacy
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Also, if we put <math>x=0</math> in the above equation we get <math>f(2) = 0</math>
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<math>f(x) = 2-x</math> satisfies both the above.
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Hence, the solutions are <math>\boxed{\color{red}{f(x) = x}}</math> and <math>\boxed{\color{red}{f(x) = 2-x}}</math>.
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==See Also==
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{{IMO box|year=2015|num-b=4|num-a=6}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Latest revision as of 23:24, 14 February 2024

Problem

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f$:$\mathbb{R}\rightarrow\mathbb{R}$ satisfying the equation

$f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$

for all real numbers $x$ and $y$.

Proposed by Dorlir Ahmeti, Albania

Solution

$f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)$ for all real numbers $x$ and $y$.

(1) Put $x=y=0$ in the equation, We get$f(0 + f(0)) + f(0) = 0 + f(0) + 0$ or $f(f(0)) = 0$ Let $f(0) = k$, then $f(k) = 0$

(2) Put $x=0, y=k$ in the equation, We get $f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)$ But $f(k) = 0$ and $f(0) = k$ so, $f(0) + f(0) = f(0)^2$ or $f(0)[f(0) - 2] = 0$ Hence $f(0) = 0, 2$

Case $1$ : $f(0) = 0$

Put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) = f(x)$ Say $f(x) = z$, we get $f(z) = z$

So, $f(x) = x$ is a solution -- fallacy

Case $2$ : $f(0) = 2$ Again put $x=0, y=x$ in the equation, We get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$ or, $f(f(x)) + 2 = f(x) + 2x$

We observe that $f(x)$ must be a polynomial of power $1$ as any other power (for that matter, any other function) will make the $LHS$ and $RHS$ of different powers and will not have any non-trivial solutions. -- fallacy

Also, if we put $x=0$ in the above equation we get $f(2) = 0$

$f(x) = 2-x$ satisfies both the above.

Hence, the solutions are $\boxed{\color{red}{f(x) = x}}$ and $\boxed{\color{red}{f(x) = 2-x}}$.

See Also

2015 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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