The sets have to both be infinite Now assume that there are three numbers in A Then the sum of any two of those three have to be in B Then the difference of any two numbers has to be in either A or B

Quote: The sets have to both be infinite Why :

at least one set has to be infinite since their union is N. If one is finite then let x = the largest of finite the other has x,x+1,2x+1

Quote: at least one set has to be infinite since their union is N. If one is finite then let x = the largest of finite the other has x,x+1,2x+1 Now it is OK my friend But if you check carefully your post you can find that "The sets have to both be infinite" is written there.

I'm clarifying what I think is vg23nov's solution, not claiming that I came up with this. vg meant that the case where one set is finite easily satisfies the condition; just choose $ a$ and $ b$ to be both larger than the largest element of the finite set (so $ a$ and $ b$ are in the infinite set) and $ a+b$ will also be in the infinite set. He also solves the case where both sets are infinite by taking three large enough numbers in $ A$ and building up a set that works by considering where the numbers' sums and differences can be.

Suppose that there exists an integer n such that no of the triples (a,b,a+b) where a,b both greater than n , belongs to either A or B In the case one of set is finite, then we can take the solution of vg23nov. If both sets are infinite, as A,B is an partition of N. There exist a number infinite integer M such that M belongs to A and M+1 belongs to B Let's take such a M with M>2.n Consider the couples (n+1, M-n-1), (n+2, M-n-2),..., (n+k, M-n-k) where k is the largest integer less than (M-2n) /2 We could see that A contains at most k elements from the set {n+1,n+2,...,M-n-1} Consider the couples (n+1, M+1-n-1), (n+2, M+1-n-2),..., (n+k, M+1-n-k) where l is the largest integer less than (M+1-2n) /2 We could see that A contains at most l elements from the set {n+1,n+2,...,M+1-n-1} Then A U B contains at most k+l elements from the set {n+1,n+2,.., M+1-n-1} , we could show that there exists an integer from this set and not belongs to neither A nor B (contradiction)

Let $a,b,c\in A,a,b,c>n$.(If there doesn't exist,replace $A$ with $B$). $a,b,c\in A\Rightarrow a+b,c+a\in B\Rightarrow 2a+b+c\in A\Rightarrow a+b+c\in B\Rightarrow 2a+2b+c\in A$ Thus $b,2a+b+c,2a+2b+c$ satisfy.