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2015 IMO Problems/Problem 5: Difference between revisions

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{{solution}}
{{solution}}
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
<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>.
Proposed by Dorlir Ahmeti, Albania
{{solution}}
All forms of a function can be expressed in a major form as
as
<math>f(r) = ar^t + c</math>
using this,  <math>f(x+f(x+y))+f(xy) = a(x + a(x+y)^t + c)^t + c +a(xy)^t + c</math>
and    <math>x+f(x+y)+yf(x) =x + a(x+y)^t + c + ay(x)^t +yc</math>
and for both expressions to be equal,
<math>t</math> has to be 1,
<math>c</math> has to be 0,
and <math>a</math> has to be 1.
therefore the function that can satisfy the equation in the question is <math>f(r) = r^1 + 0</math>
which is <math>f(r) = r</math>.


[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Algebra Problems]]
[[Category:Functional Equation Problems]]
[[Category:Functional Equation Problems]]

Revision as of 17:43, 9 October 2016

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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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

This problem needs a solution. If you have a solution for it, please help us out by adding it. All forms of a function can be expressed in a major form as as $f(r) = ar^t + c$ using this, $f(x+f(x+y))+f(xy) = a(x + a(x+y)^t + c)^t + c +a(xy)^t + c$ and $x+f(x+y)+yf(x) =x + a(x+y)^t + c + ay(x)^t +yc$ and for both expressions to be equal, $t$ has to be 1, $c$ has to be 0, and $a$ has to be 1. therefore the function that can satisfy the equation in the question is $f(r) = r^1 + 0$ which is $f(r) = r$.

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