Difference between revisions of "2012 IMO Problems/Problem 4"
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<cmath>f(n\cdot{a}) = n^2f(a)</cmath> | <cmath>f(n\cdot{a}) = n^2f(a)</cmath> | ||
We note that <math>f(a)=a^2f(1).</math> This means that we want to find what the possible values of <math>f(1)</math> are in order to finish the problem. <math>a + b + c = 0 \Rightarrow</math> <math>c = -(a+b) \Rightarrow</math> <math>a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =</math> <math>2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.</math> Returning to the above manipulations (the ones used to show that <math>f(n\cdot{a}) = n^2f(a)</math>), we see that letting <math>n = \frac{a}{b}</math> and multiplying through by <math>b^4</math> yields precisely this result (again, assuming that <math>f(1) \neq 0,</math> with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, <math>f(1)</math> can be any integer value (since <math>f</math> maps the integers to the integers only), and setting <math>f(1) = m</math>, we see that any function of the form <cmath>f(a) = m\cdot{a^2}, m \in \mathbb{Z}</cmath> satisfies the condition. | We note that <math>f(a)=a^2f(1).</math> This means that we want to find what the possible values of <math>f(1)</math> are in order to finish the problem. <math>a + b + c = 0 \Rightarrow</math> <math>c = -(a+b) \Rightarrow</math> <math>a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =</math> <math>2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.</math> Returning to the above manipulations (the ones used to show that <math>f(n\cdot{a}) = n^2f(a)</math>), we see that letting <math>n = \frac{a}{b}</math> and multiplying through by <math>b^4</math> yields precisely this result (again, assuming that <math>f(1) \neq 0,</math> with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, <math>f(1)</math> can be any integer value (since <math>f</math> maps the integers to the integers only), and setting <math>f(1) = m</math>, we see that any function of the form <cmath>f(a) = m\cdot{a^2}, m \in \mathbb{Z}</cmath> satisfies the condition. | ||
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-Gadmaget | -Gadmaget |
Revision as of 21:04, 14 June 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
What about
Then
We conjecture that
Consider
and assume that
If it does, we get that the constant 0 function satisfies the conditions of the problem.
We note that
This means that we want to find what the possible values of
are in order to finish the problem.
Returning to the above manipulations (the ones used to show that
), we see that letting
and multiplying through by
yields precisely this result (again, assuming that
with equality yielding the fact that the constant 0 function satisfies the condition). Therefore,
can be any integer value (since
maps the integers to the integers only), and setting
, we see that any function of the form
satisfies the condition.
-Gadmaget