Difference between revisions of "2015 IMO Problems/Problem 5"
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(2) Put <math>x=0, y=k</math> in the equation, | (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> | 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> | + | But <math>f(k) = 0</math> and <math>f(0) = k</math> |
so, <math>f(0) + f(0) = f(0)^2</math> | so, <math>f(0) + f(0) = f(0)^2</math> | ||
or <math>f(0)[f(0) - 2] = 0</math> | or <math>f(0)[f(0) - 2] = 0</math> | ||
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Say <math>f(x) = z</math>, we get <math>f(z) = z</math> | Say <math>f(x) = z</math>, we get <math>f(z) = z</math> | ||
− | So, <math>f(x) = x</math> is a solution | + | So, <math>f(x) = x</math> is a solution -- fallacy |
Case <math>2</math> : <math>f(0) = 2</math> | Case <math>2</math> : <math>f(0) = 2</math> | ||
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or, <math>f(f(x)) + 2 = f(x) + 2x</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. | + | 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> | Also, if we put <math>x=0</math> in the above equation we get <math>f(2) = 0</math> | ||
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Hence, the solutions are <math>\boxed{\color{red}{f(x) = x}}</math> and <math>\boxed{\color{red}{f(x) = 2-x}}</math>. | Hence, the solutions are <math>\boxed{\color{red}{f(x) = x}}</math> and <math>\boxed{\color{red}{f(x) = 2-x}}</math>. | ||
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==See Also== | ==See Also== |
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 |