# A discrete maths problem on $2$ sets

We have $2$ sets, $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$ with $a_i, b_j\in\mathbb{Z}$.

Is there any condition on elements of $A$ and $B$ that yield $A+B=\{a_i+b_j| i,j \in\{1,2,3\}\}$ or $A.B=\{a_i.b_j| i,j \in\{1,2,3\}\}$ have $3$ distinct elements with any multiplicities?

$A+B$ must have at least four members. The elements $a_1+b_1, a_1+b_2,$ and $a_1+b_3$ are all distinct. $a_2+b_1, a_2+b_2$ and $a_2+b_3$ cannot be a permutation of the first three, as the first three sum to $3a_1+b_1+b_2+b_3$ which must be different from $3a_2+b_1+b_2+b_3$, which is the sum of the second three. $A=B=\{0,1,2\}$ shows we can get to five. Five is the minimum as Fan Zheng shows below.

• WLOG suppose $a_1<a_2<a_3$ and $b_1<b_2<b_3$. Then $a_1+b_1<a_1+b_2<a_1+b_3<a_2+b_3<a_3+b_3$, so $|A+B|\ge5$. Commented Sep 27, 2016 at 17:14
• Sorry for the downvote, but it is not the same for products, because multiplication by nonpositive numbers does not preserve the ordering. In fact, $|A.B|$ can be 3. This happens when $A=B=\{-1,0,1\}$ and $A.B$ is then also $\{-1,0,1\}$. Commented Sep 27, 2016 at 17:17
• @FanZheng: Good points. Thanks. Commented Sep 27, 2016 at 19:09