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> | ||
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+ | <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 f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math> | <math>\Rightarrow 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 |
Revision as of 05:46, 4 December 2013
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.
So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.
--Dineshram