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If $A,B\subseteq\{0,1,2,\ldots\},$ then $A+B=\{a+b:a\in A,b\in B\}.$ Is it true that for any $X\subseteq \{0,1,2,\ldots\}$ infinite, there exist infinite sets $A,B\subseteq\{0,1,2,\ldots\}$ such that $A+B=X?$

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If $b_1 \ne b_2 \in B$, there are infinitely many pairs of elements of $A + B$ that differ by $b_2 - b_1$. But for example $\{1,2,4,8,\ldots\}$ has all its differences distinct.

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The set $P$ of all primes cannot be written as a sum of two infinite sets.

Suppose $A=\{a_0,a_1,\ldots\},$ $a_0<a_1<\ldots$ and $B=\{b_0,b_1,\ldots\}$, $b_0<b_1<\ldots$ are two infinite sets of natural numbers such that $A+B=P$. We can assume $a_0\geq b_0.$ Then either $a_0=2$ and $b_0=0$, or $a_0=b_0=1.$ In the first case, $a_1$ must be a prime in order for $a_1+0$ to be a prime. But then $b_1>0$ must be even in order for $a_1+b_1$ to be odd. But then $a_0+b_1>2$ is even and so not a prime.

The other case is just as simple.

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