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 | ||
− | + | ==Solution== | |
− | + | ||
+ | <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</math> and <math>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, | ||
+ | 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> | ||
+ | |||
+ | So, <math>f(x) = x</math> is a solution -- fallacy | ||
+ | |||
+ | 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. -- fallacy | ||
+ | |||
+ | Also, if we put <math>x=0</math> in the above equation we get <math>f(2) = 0</math> | ||
− | <math>f(x | + | <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>. | |
− | + | ==See Also== | |
− | {{ | + | {{IMO box|year=2015|num-b=4|num-a=6}} |
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Functional Equation Problems]] | [[Category:Functional Equation Problems]] |
Latest revision as of 00:24, 15 February 2024
Problem
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
and
so,
or
Hence
Case :
Put in the equation,
We get
or,
Say
, we get
So, is a solution -- fallacy
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. -- fallacy
Also, if we put in the above equation we get
satisfies both the above.
Hence, the solutions are and
.
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 |