Difference between revisions of "2012 IMO Problems/Problem 4"
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Now we look at <math>b = -a, c = 0.</math> <math>f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow</math> <math>f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow</math> <math>f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow</math> <math>(f(a) - f(-a))^2=0 \Rightarrow</math> <cmath>f(a)=f(-a).</cmath> | Now we look at <math>b = -a, c = 0.</math> <math>f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow</math> <math>f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow</math> <math>f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow</math> <math>(f(a) - f(-a))^2=0 \Rightarrow</math> <cmath>f(a)=f(-a).</cmath> | ||
− | We can write <math>f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0 \Rightarrow</math> <cmath>f(c) = f(-c) = f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}</cmath> | + | We can write <math>f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0 \Rightarrow</math> |
+ | |||
+ | <cmath>f(c) = f(-c) = f(a+b) =\frac{2(f(a)+f(b)) \pm \sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}</cmath> | ||
+ | |||
+ | <cmath>\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}</cmath> | ||
If <math>f(b) = 0</math>, then <cmath>f(a+b) = f(a) = f((a)mod(b))</cmath> | If <math>f(b) = 0</math>, then <cmath>f(a+b) = f(a) = f((a)mod(b))</cmath> | ||
− | '''Case 1''': <math>f(1) = 0 \Rightarrow f(x)= 0</math> <math>\forall</math> <math>x</math> <math>\Box</math> | + | '''Case 1''': <math>f(1) = 0 \Rightarrow f(x)= 0</math> <math>\forall</math> <math>x.</math> <math>\Box</math> |
Case 2: <math>f(1) \not= 0</math>, we will have <math>f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0</math> or <math>f(2) = 4f(1)</math> | Case 2: <math>f(1) \not= 0</math>, we will have <math>f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0</math> or <math>f(2) = 4f(1)</math> | ||
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<math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well. | <math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well. | ||
− | <math> | + | As the statement is true for <math>m = 4</math>, by mathematical induction we can conclude |
+ | |||
+ | <math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math> | ||
+ | |||
+ | '''So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.''' | ||
--Dineshram | --Dineshram | ||
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==See Also== | ==See Also== | ||
*[[IMO Problems and Solutions]] | *[[IMO Problems and Solutions]] | ||
+ | |||
+ | {{IMO box|year=2012|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 07:08, 20 November 2023
Find all functions such that, for all integers and that satisfy , the following equality holds: (Here denotes the set of integers.)
Solution
Consider Then Now we look at
We can write
If , then
Case 1:
Case 2: , we will have or
Case 2.1: if is odd, if is even.
Case 2.2:
or
Case 2.2.1:
and
or and or
Case 2.2.2:
or
and or
If then
We will prove by induction
If then is true for some .
and if the statement is true for
or
and or
the statement is true for as well.
As the statement is true for , by mathematical induction we can conclude
So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.
--Dineshram
See Also
2012 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |