$A+A\subseteq A\times A$ Define $A+A=\{a+b\colon a,b\in A\}$,$A\times A =\{ab\colon a,b\in A\}$.
Does there exist a finite integer set $A\subseteq \mathbb{Z}^+$, such that $|A|>1$ and $A+A\subseteq A\times A$ ?

We assume such a set exists and derive a contradiction. First, if $\{1,2\} \subset A$ then $3 \in A$ as $1+2=3$ so we must have $1 \cdot 3 \in A \times A$. Also, since $|A| > 1$ then if $1$ or $2$ is not in $A$ then there exist an element of $A$ greater than $2$, so in either case such an element exists.
Edit: the rest is wrong so I struck it out. 
Let $x$ be the smallest such element. Now as 2x \in A + A we must have 2 \in A because we need 2x \in A \times A and the minimality of x. This in turn implies implies x + 2 \in A \times A.
So 1,2,4,x,2x and x^2 are the smallest possible elements of A \times A and because of our choice of x we notes 2+x < 2x < x^2 so 2+x must be on that list. However proceeding by cases we see that x+2=1, x+2 = 2, x+2=4, x+2=x and x+2=2x all contradict our choice of x so no such set A can exist.
