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Imagine that the integers are split into sets $A$, $B$, $C$ with the restriction that the negative of an integer belonging to $A$ should belong to $B$, and ineteger that can be represented as the sum of two integers belonging to $B$ should be in $A$. It is fairly simple to show that the converse of both these restrictions should also hold true- the negative of an integer in $B$ is in $A$ and the sum of two integers in $A$ is in $B$.

In what sorts of ways can we arrange all integers in these sets?

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Your example doesn't work, as the sums of elements of $B$ (positive integers) aren't in $A$ (negative integers). –  jwodder Apr 6 '12 at 0:44
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I interpret "split" as meaning $A$, $B$, $C$ must be disjoint and nonempty. Let $a_0$ be the member of $A$ with least absolute value. Then for integers $k$, $k a_0 \in A$ if $k \equiv 1 \mod 3$, $B$ if $k \equiv 2 \mod 3$, $C$ if $k \equiv 0 \mod 3$.
Moreover, every integer not a multiple of $a_0$ must be in $C$.

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Could you explain why this is further? Also are the roles of A and B effectively interchangeable in your solution? –  Ali Apr 13 '12 at 15:58
    
Yes, the problem is symmetric in $A$ and $B$, and $B = -A$. –  Robert Israel Apr 15 '12 at 5:51
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