Difference between revisions of "2015 IMO Problems/Problem 5"
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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>. | <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>. |
Revision as of 12:06, 14 May 2017
Let be the set of real numbers. Determine all functions : satisfying the equation
for all real numbers and .
Proposed by Dorlir Ahmeti, Albania
Solution:
for all real numbers and .
(1) Put in the equation, We get or Let , then
(2) Put in the equation, We get But so, or Hence
Case :
Put in the equation, We get or, Say , we get
So, is a solution
Case : Again put in the equation, We get or,
We observe that must be a polynomial of power as any other power (for that matter, any other function) will make the and of different powers and will not have any non-trivial solutions.
Also, if we put in the above equation we get
satisfies both the above.
Hence, the solutions are and .