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Difference between revisions of "2015 IMO Problems/Problem 5"
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f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)
+
<math>f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)</math>
  
−
(1) Put x=y=0 in the equation,
+
(1) Put <math>x=y=0</math> in the equation,
−
We get f(0 + f(0)) + f(0) = 0 + f(0) + 0
+
We get<math> f(0 + f(0)) + f(0) = 0 + f(0) + 0
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or f(f(0)) = 0
+
or f(f(0)) = 0</math>
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Let f(0) = k, then f(k) = 0
+
Let <math>f(0) = k</math>, then <math>f(k) = 0</math>
  
−
(2) Put x=0, y=k in the equation,
+
(2) Put <math>x=0, y=k</math> in the equation,
−
We get f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)
+
We get <math>f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)</math>
−
But f(k) = 0 and f(0) = k
+
But <math>f(k) = 0 and f(0) = k</math>
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so, f(0) + f(0) = f(0)^2
+
so, <math>f(0) + f(0) = f(0)^2</math>
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or f(0)[f(0) - 2] = 0
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or <math>f(0)[f(0) - 2] = 0</math>
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Hence f(0) = 0, 2
+
Hence <math>f(0) = 0, 2</math>
  
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Case I : f(0) = 0
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Case <math>1</math> : <math>f(0) = 0</math>
  
−
Put x=0, y=x in the equation,
+
Put <math>x=0, y=x</math> in the equation,
−
We get f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)
+
We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
−
or, f(f(x)) = f(x)
+
or, <math>f(f(x)) = f(x)</math>
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Say f(x) = z, we get f(z) = z
+
Say <math>f(x) = z</math>, we get <math>f(z) = z</math>
  
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So, f(x) = x is a solution
+
So, <math>f(x) = x</math> is a solution
  
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Case II : f(0) = 2
+
Case <math>2</math> : <math>f(0) = 2</math>
−
Again put x=0, y=x in the equation,
+
Again put <math>x=0, y=x</math> in the equation,
−
We get f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)
+
We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
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or, f(f(x)) + 2 = f(x) + 2x
+
or, <math>f(f(x)) + 2 = f(x) + 2x</math>
  
−
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.
+
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.
  
−
Also, if we put x=0 in the above equation we get f(2) = 0
+
Also, if we put <math>x=0</math> in the above equation we get <math>f(2) = 0</math>
  
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f(x) = 2-x satisfies both the above.
+
<math>f(x) = 2-x</math> satisfies both the above.
  
−
Hence, the solutions are f(x) = x and f(x) = 2-x.
+
Hence, the solutions are <math>\boxed{\color{red}{f(x) = x}}</math> and <math>\boxed{\color{red}{f(x) = 2-x}}</math>.
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Functional Equation Problems]]
 
[[Category:Functional Equation Problems]]

Revision as of 12:03, 14 May 2017

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.


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

(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

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.

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 $\boxed{\color{red}{f(x) = 2-x}}$.

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