If a and b are any real numbers we have:

I. If a@b=0 then ab-b=0, which is the case when b=0 or a=1

II. If (a+b)@b=0 then (a+b)b-b=0, which is the case when b=0 or (a+b)=1

III. If a@(a+b)=0 then a(a+b)-(a+b)=0 which is the case when (a+b)=0 or a=1.

Now if instead a and b are positive integers I. reduces to a=1, II. has no possible solutions in positive integers (explain why), while III. again permits a=1.

RonL