Find all sets $A$ and $B$ that satisfy the following conditions: a) $A \cup B= \mathbb{Z}$; b) if $x \in A$ then $x-1 \in B$; c) if $x,y \in B$ then $x+y \in A$. Laurentiu Panaitopol
Problem
Source: Romanian IMO Team Selection Test TST 2002, problem 1
Tags: induction, number theory unsolved, number theory
socrates
04.07.2005 16:54
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grobber
04.07.2005 16:57
I believe it has been discussed before on the forum. It was used in the first Romanian TST in 2002, if I am not mistaken.
ifai
19.10.2005 16:09
socrates wrote: A is the set of the even and B of the odd numbers! $A=B=\mathbb{Z}$ is also the answer.
Kondr
20.10.2005 21:35
If 0 was in B, by c) 0 is also in A, so 0 is in A. By b) -1 is in B, by c) -2 is in A, by c) -3 is in B. By induction if -2k+1 is in B, by c) -2k is in A and -2k-1 is in B, so all negative odd numbers are in B. By b) , all negative even numbers are in A. If 1 iwas not in B, 1 would be in A, by b) 0 would be in B, so by c) if x is in B, x is also in A, so all numbers would be in A. Because of that all numbers would be in B to, contradiction. So 1 is in B. By c) 2 is in b. By induction it's simple to show that all positive odds are in B and all positiv evens in A. If there were no other numbers in the sets, the solution wolud be set of evens and set of odds. If there was an odd number in A, there would be by b) an even number in A. So in any other solution then set of evens and set of odds there must be an even number in B. But then 1 is in A, 0 in B so all the numbers from B are in A too and by b) all are in A as well. So the only one solutions are (evens,odds), ($\mathbb Z$,$\mathbb Z$).
WakeUp
05.02.2011 18:50
See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=30568