$A+A\subseteq A\times A$ 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 $A+A\subseteq A\times A$ ?
 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.
A: The answer is no EDIT so far all cases work except case 3
case 1: let assume 1,2 ∉ A
Now let a∈A where a is the smallest valid option.
so a+a ∈ A + A
so the smallest value in A x A is 2a (because a was the smallest number in A),
but a+a < a + a + a .... + a (a lots of a) = a*a (as a >2)
so a+b ∉ A x A
so A+A ⊊ A×A
case 2: 1 ∈ A, 2 ∉ A
2 = 1+1 ∈ A + A but 2 ∉ A X A
case 3: 1,2 ∈ A EDIT this argument doesn't work sorry!
let p>2 be the smallest positive prime integer not in A (A is finite so p must exist).
so (p-1) ∈ A
p = 1+(p-1) ∈ A + A either but p ∉ A X A, because the only way to make p= 1*p but p ∉ A
case 4: 1 ∉ A, 2 ∈ A:
let b>2 be the smallest positive integer in A other than 2.
then 2+b ∈ A + A but 2+b ∉ A X A, otherwise 2+b=cd where c,d ∈ A and so cannot be 1, so both c,d ≥ 2 => 
2 ≤ c ≤ (2+b)/2 < b
and 
2 ≤ d ≤ (2+b)/2 < b
so if either c or d are not equal to 2 say c then c ∉ A because its smaller than b which contradicts the first statement in this case, so 2+b ∉ A X A. 
Finally c = d = 2 but in that case we now have 2 + b = 4 which meant b=2 which contradicts the assumption that b > 2.
so all cases covered! the answer is no EDIT so far all cases work except case 3
