Difference between revisions of "2015 IMO Problems/Problem 5"
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Proposed by Dorlir Ahmeti, Albania | Proposed by Dorlir Ahmeti, Albania | ||
− | + | <math>\bold{Solution}</math>: | |
− | |||
− | <math>f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)</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>. |
− | + | (1) Put <math>x=y=0</math> in the equation, | |
+ | We get<math> f(0 + f(0)) + f(0) = 0 + f(0) + 0</math> | ||
+ | or <math>f(f(0)) = 0</math> | ||
+ | Let <math>f(0) = k</math>, then <math>f(k) = 0</math> | ||
− | + | (2) Put <math>x=0, y=k</math> in the equation, | |
+ | We get <math>f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)</math> | ||
+ | But <math>f(k) = 0 and f(0) = k</math> | ||
+ | so, <math>f(0) + f(0) = f(0)^2</math> | ||
+ | or <math>f(0)[f(0) - 2] = 0</math> | ||
+ | Hence <math>f(0) = 0, 2</math> | ||
− | + | Case <math>1</math> : <math>f(0) = 0</math> | |
− | + | ||
− | + | Put <math>x=0, y=x</math> in the equation, | |
− | <math>f( | + | We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math> |
− | + | or, <math>f(f(x)) = f(x)</math> | |
− | + | Say <math>f(x) = z</math>, we get <math>f(z) = z</math> | |
− | + | ||
− | <math> | + | So, <math>f(x) = x</math> is a solution |
− | <math> | + | |
− | and <math> | + | Case <math>2</math> : <math>f(0) = 2</math> |
− | + | Again put <math>x=0, y=x</math> in the equation, | |
− | + | We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math> | |
+ | or, <math>f(f(x)) + 2 = f(x) + 2x</math> | ||
+ | |||
+ | 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 <math>x=0</math> in the above equation we get <math>f(2) = 0</math> | ||
+ | |||
+ | <math>f(x) = 2-x</math> satisfies both the above. | ||
+ | |||
+ | 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 13:08, 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
:
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 .