A troublesome logic puzzle. 
Here're two distinct integers $x,y$, both greater than $1$ but less than $100$. Student $A$ and $B$ know the values of $xy$ and $x+y$ respectively. Let's hear their conversation.
$A$: I don't know what $x,y$ are.
$B$: Me too, but I already know that you don't know $x,y$ before you told me.
$A$: I know what $x,y$ are now!
$B$: Me too!
Assuming $A$ knows $B$ knows $x+y$, $B$ knows $A$ knows $xy$, $B$ knows $A$ knows $B$ knows $x+y$, $A$ knows $B$ knows $A$ knows $xy$... bla bla bla, what is $x$ and $y$?

I've only figure out that $x,y$ can't be prime. Can someone please help me? Thanks.
 A: Hint:
Let's look at the second statement.
What must be true about B's sum ($x+y$) for him to make such a statement? Every pair of numbers that sums to $x+y$ must contain at least one composite number (which would make their product impossible to deconstruct, so to speak). 
For example, $(17)=2+15=3+14=4+13=5+12=6+11=7+10=8+9$, so $17$ is such a number.
Then the third statement tell us that of all of the factor pairs of A's number ($xy$), only one pair sums up to "such a number" (as $17$). For example, A could have $60 =2 \cdot 30$, since $2+30=3+29$ is not "such a number". But we'd have to check every factor pair of $60$... 
If my mental calculations are correct, the sums of (proper) factor pairs of $60$ are $32,23,19,17,16$... and only $23$ and $17$ are "such numbers". But we are looking for a product whose factor pairs have only ONE such number amongst their sums.
Can you take it from here?
A: UPDATE: The solution is $(x,y) = (4,13)$.
I have reworked the post below. I show that this solution satisfies the evolution of the puzzle, but I have not proved it in abstract, nor did I proved uniqueness. Any comment, addition, or correction, is welcome.
Fact collection
According to the premises, the set of common knowledge $K$ before the conversation is:  


*

*$x,y$ are integers, distinct ($x\neq y$), greater than $1$ and smaller than $100$.  

*$A$ knows privately $P=xy$ and $B$ knows privately $S=x+y$.


During the conversation the set of common knowledge is augmented by  $\Delta$:


*

*$A$ did not know $(x,y)$.

*$B$ did not know $(x,y)$.

*$B$ knew that $A$ could not know $(x,y)$.


...And we are told that $K\cup \Delta \cup P $ permits $A$ to deduce $(x,y)$. And sequentially, given that $A$ deduced $(x,y)$ and declared so, $B$ was able to deduce it too.
Analysis
Since $A$ found the answer with fewer information than $B$ , we must first examine her. Before the conversation, $A$ knew $K$, and the product, $P$. The only thing she could do was to factorize the product into primes, $P = p_1p_2...p_n$, calculate all possible sub-products in groupings of two, take those that gave pairs of numbers permitted by $K$, and form a "possible solutions" set, let's denote it $A|(K\cup P) = \{(x_1,y_1),...,(x_m,y_m)\}$.
Then it must be the case that with the information $A$ learned during the conversation, she was able to reduce $A|(K\cup P)$ into a singleton, write $A|(K\cup \Delta\cup P)=A^*$. What did she learn?  
1) That $B$ was certain that $P$ was such that could not lead $A$ to the answer. How $B$ could be certain about that? The only way that $A$ could find the answer by knowing $K$ and only $P$ privately, would be if $P$ was factored into only two (and permissible by $K$) primes (counting multiplicities of $2$ as separate factors). So $B$ must have been certain that the product $P$ was not factored in only two primes. How? Primes are odd numbers (except $2$), and the sum of two odd numbers is even. So a necessary condition for $B$ to be certain about $P$ not providing $A$ with enough information, is that the sum $S$ is odd. Still, an odd number can always be written as $2+odd$. But if "odd" was also prime, $A$ would knew before hand. So the information that $B$ conveyed to $A$ by the relevant statement is :
"$S$ is an odd number, and the solution does not include two primes". But only the first part is new information to $A$.  
2) The other thing that $A$ learned from $B$, is that $B$ did not know the answer. But I could not find a use for this information, so I ignore it (maybe this is problematic?)
So: if learning that $S$ is $odd$ was enough for $A$ to deduce the solution, then it must be the case that the set of possible solutions $A|(K\cup P)$ before the conversation, contained only one pair of numbers the sum of which was odd, while all other pairs had numbers the sum of which was even. In all cases, for a prime factorization to lead to groupings of two composite factors having some even and some an odd sum, it must be that $2$ appears at least (or only, I am not sure) twice as a factor. (For example the number $30= 2 \times 3 \times 5$. The possible groupings in two are $(2,15), (6, 5), (3, 10)$. The sum of each pair is an odd number). 
Solution 
We show now that the pair $(4,13)$ conforms with the evolution of the puzzle. 
Deduction of the solution by $A$ 
The product of the two numbers is $P=52$. The prime factorization of $52 = 2\times 2\times 13$. Then the set of possible solutions of $A$ before the conversation was $A|(K\cup P=52)=\{(4,13), (2,26)\}$. Once $A$ realized through the conversation that the sum of the two numbers is odd, she could exclude the pair $(2,26)$, and declare that she knew now the solution: $(4,13)$ (without revealing it of course).
Deduction of the solution by $B$ 
If the pair was indeed $(4,13)$, then $B$ knew that $S=17$. There are seven pairs of two numbers greater than one that can add up to $17$. I list them below writing their product and their prime factorization
\begin{array}{cccc}
Pair & Product & Factorization & \text {Sums of factor groupings} \\
(2, 15)& 30 & 2\times 3\times 5 & \text{all odd} \\
(3, 14) & 42 & 2\times 3\times 7 & \text{all odd} \\
(4,13) & 52 & 2\times 2\times 13 & \text{one odd, one even} \\
(5,12) & 60 & 2\times 2\times 3 \times 5 & \text{more than one odd} \\
(6,11) & 66 & 2\times 3\times 11 & \text{all odd} \\
(7,10) & 70 & 2\times 5\times 7 & \text{all odd} \\
(8,9) & 72 & 2\times 2\times 2\times 3 \times 3 & \text{more than one odd} \\
\end{array} 
The above is what $B$ knows. Hearing that $A$ deduced the solution by only learning that the sum of the pair was odd, he can in turn deduce that only the pair $(4,13)$ would permit $A$ to find the solution. So $B$ too found it.
