TITLE: an arithmetic sum and product puzzle
QUESTION [6 upvotes]: Assume that $X$ and $Y$ are positive integers with $1<X<Y$. Mr. S knows the value $X+Y$, and Mr. P knows the value $XY$. They then have the following conversation.
Mr. S says to Mr. P : "You do not know the value of $X$ and $Y$"
Mr. P responds to Mr. S: "Now that you said this to me, I know the values of $X$ and $Y$."
Mr. S then responds : "So do I".
Find $X$ and $Y$.
My (very inadequate!) thoughts on this puzzle so far are as follows. It is clear that $X$ and $Y$ cannot be two distinct primes-else P would know their values instantly. I also thought (assuming Goldbach's conjecture (is this legit?!) that $X+Y$ cannot be even. Because if $X+Y$ were even, Mr. S could not rule out the possibility that $XY$ is the product of two distinct primes. Thus, all even numbers in $[1,Y]$ can be ruled out. The same reasoning also rules out odd positive integers that are of the form $2$ added to an odd prime, and squares of primes in $[1,Y]$.
The problem as stated did not give a numerical bound on $Y$. If it had-like (say) $Y=100$, I would be tempted to try cases and see which integers can be ruled in-many seem to be ruled out, and there could well be more than one solution. Is there a general argument that works for any positive integer $Y$ that I am missing?

REPLY [2 votes]: A solution to this problem is $X=4,Y=13$.
Proof this works
S knows that $(X,Y)$ is one of $(2,15)(3,14),(4,13),(5,12),(6,11),(7,10),(8,9)$. The product $XY$ is then one of $30,42,52,60,66,70,72$. Each of these has at least two permissible expressions as a product and so P does not initially know the two values. 
P knows that $(X,Y)$ is one of $(2,26),(4,13)$. However, if $(X,Y)=(2,26)$, then S could not exclude the possibility $(X,Y)=(5,23)$ when P would initially know the two values. So, once S has spoken, P  knows that $(X,Y)=(4,13)$. 
Now back to S who knows that P has been able to find the values.  For each possible $XY$ other than $52$ he can see that P would not be able to distinguish between  the solution with $X+Y=17$ and :- 
$XY=30$. $X=5,Y=6.$
$XY=42$. $X=2,Y=21.$
$XY=60$. $X=3,Y=20.$
$XY=66$. $X=2,Y=33.$
$XY=70$. $X=2,Y=35.$
$XY=72$. $X=3,Y=24.$
Proving that there are no other solutions looks extremely messy!