TITLE: Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z^2$
QUESTION [15 upvotes]: Is there a torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq  A \oplus \Bbb Z  \oplus \Bbb Z$ (as groups)?

Notice that $\Bbb Z$ is not cancellable, so 
$A \oplus \Bbb Z \simeq  (A \oplus \Bbb Z)  \oplus \Bbb Z$ doesn't imply that $A \simeq A \oplus \Bbb Z$. Combined with this question, such a group $A$ would possibly provide an answer to that question.
An example of torsion free abelian group $A$ such that $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$ was given there.
The example was the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$, i.e. 
$$A = 
\left\{
(x_n)_{n \geq 1} \subset \Bbb Z[\sqrt 2]          \;\;:\;\;
\exists C>0,\;  \forall n \geq 1,\;  |x_n| \leq C
\right\}$$
I wasn't able to adapt this example in order to answer my question.

REPLY [17 votes]: $\mathbb{Z}$ is cancellable for abelian groups. This was proved in the 1950s by Walker and Cohn (independently) and is often called "Walker's cancellation theorem". The proof is only a few lines.
So if $A$ is an abelian group with $A\oplus\mathbb{Z}\cong A\oplus\mathbb{Z}^2$, then $A\cong A\oplus\mathbb{Z}$.