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Difference between revisions of "2012 IMO Problems/Problem 4"
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(Here <math>\mathbb{Z}</math> denotes the set of integers.)
 
(Here <math>\mathbb{Z}</math> denotes the set of integers.)
  
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'''Solution:'''
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==Solution==
 
Consider <math>a = b = c = 0.</math> Then <math>f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow</math> <cmath>f(0) = 0.</cmath>
 
Consider <math>a = b = c = 0.</math> Then <math>f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow</math> <cmath>f(0) = 0.</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>
 
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>

Revision as of 22:57, 10 October 2013

Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$, the following equality holds: \[f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\] (Here $\mathbb{Z}$ denotes the set of integers.)

Solution

Consider $a = b = c = 0.$ Then $f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow$ \[f(0) = 0.\] Now we look at $b = -a, c = 0.$ $f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow$ $f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow$ $f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow$ $(f(a) - f(-a))^2=0 \Rightarrow$ \[f(a)=f(-a).\] What about $b = a, c = -2a?$ Then $f(a)^2 + f(a)^2 + f(-2a)^2 = 2f(a)f(a) + 2f(a)f(-2a) + 2f(-2a)f(a) \Rightarrow$ $2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a) \Rightarrow$ $f(-2a)^2 = 4f(a)f(-2a) \Rightarrow$ \[f(-2a) = f(2a) = 4f(a).\] We conjecture that $f(na) = n^2f(a).$ Consider $b = n\cdot{a}, c = -(n+1)\cdot{a}$ and assume that $f(a) \neq 0.$ If it does, we get that the constant 0 function satisfies the conditions of the problem. $f(a)^2 + f(n\cdot{a})^2 + f((n+1)\cdot{a})^2 =$ $2f(a)f(n\cdot{a}) + 2f(n\cdot{a})f((n+1)\cdot{a}) + 2f((n+1)\cdot{a})f(a) \Rightarrow$ $f(a)^2 + (n^2f(a))^2 + ((n+1)^2f(a))^2 =$ $2n^2f(a)^2 + 2n^2(n+1)^2f(a)^2 + 2(n+1)^2f(a)^2 \Rightarrow$ $1 + n^4 + (n+1)^4 = 2n^2 + 2n^2(n+1)^2 + 2(n+1)^2 \Rightarrow$ $1 + n^4 + n^4 + 4n^3 + 6n^2 + 4n + 1 = 2n^2 + 2n^4 + 4n^3 + 2n^2 + 2n^2 + 4n +2 \Rightarrow$ $2n^4 + 4n^3 + 6n^2 + 4n +2 = 2n^4 + 4n^3 + 6n^2 + 4n + 2 \hspace{7 mm} \checkmark$ \[f(n\cdot{a}) = n^2f(a)\] We note that $f(a)=a^2f(1).$ This means that we want to find what the possible values of $f(1)$ are in order to finish the problem. $a + b + c = 0 \Rightarrow$ $c = -(a+b) \Rightarrow$ $a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =$ $2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.$ Returning to the above manipulations (the ones used to show that $f(n\cdot{a}) = n^2f(a)$), we see that letting $n = \frac{a}{b}$ and multiplying through by $b^4$ yields precisely this result (again, assuming that $f(1) \neq 0,$ with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, $f(1)$ can be any integer value (since $f$ maps the integers to the integers only), and setting $f(1) = m$, we see that any function of the form \[f(a) = m\cdot{a^2}, m \in \mathbb{Z}\] satisfies the condition.

-Gadmaget

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