Difference between revisions of "2012 IMO Problems/Problem 4"
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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> | ||
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− | - | + | 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> |
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+ | 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 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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+ | '''Case 2.1''': <math>f(1) \not= 0, f(2) = 0 \Rightarrow f(x) = f(x</math> <math>mod</math> <math>2) \Rightarrow f(x) = f(1)</math> if <math>x</math> is odd, <math>f(x) = 0</math> if <math>x</math> is even. <math>\Box</math> | ||
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+ | Case 2.2: <math>f(1) \not= 0, f(2) = 4f(1) \Rightarrow f(3) = f(2) + f(1) \pm 2\sqrt{f(2)f(1)}</math> | ||
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+ | <math> \Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)</math> or <math>9f(1)</math> | ||
+ | |||
+ | '''Case 2.2.1''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).</math> | ||
+ | |||
+ | <math>\Rightarrow f(4) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)}</math> and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)}</math> | ||
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+ | <math>\Rightarrow f(4) = f(1)</math> or <math>0</math> and <math>f(4) = 16f(1)</math> or <math>0</math> | ||
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+ | <math>\Rightarrow f(4) = 0 \Rightarrow f(x) = f(x</math> <math>mod</math> <math>4).\Box</math> | ||
+ | |||
+ | '''Case 2.2.2''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).</math> | ||
+ | |||
+ | <math>\Rightarrow f(4) = f(1 + 3) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)} = 16f(1)</math> or <math>4f(1)</math> | ||
+ | |||
+ | and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)} = 16f(1)</math> or <math>0. \Rightarrow f(4) = 16f(1).</math> | ||
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+ | If <math>x \le 4</math> then <math>f(x) = f(1)x^2.</math> | ||
+ | |||
+ | We will prove by induction <math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.</math> | ||
+ | |||
+ | If <math>x \le m</math> then <math>f(x) = f(1)x^2.</math> <math>\forall</math> <math>x</math> is true for some <math>m</math>. | ||
+ | |||
+ | and if the statement is true for <math>m=k</math> | ||
+ | |||
+ | <math>\Rightarrow f(k+1) = f(k) + f(1) \pm 2\sqrt{f(k)f(1)} = f(1)(k+1)^2</math> or <math>f(1)(k-1)^2</math> | ||
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+ | and <math>f(k+1) = f(k-1) + f(2) \pm 2\sqrt{f(k-1)f(2)} = f(1)(k+1)^2</math> or <math>f(1)(k-3)^2</math> | ||
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+ | <math>\Rightarrow f(k+1) = f(1)(k+1)^2.</math> | ||
+ | |||
+ | <math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well. | ||
+ | |||
+ | <math>\Rightarrow f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math> | ||
==See Also== | ==See Also== | ||
*[[IMO Problems and Solutions]] | *[[IMO Problems and Solutions]] |
Revision as of 06:30, 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.