In my opinion it’s 25 and 50
As it was said, If the product guy don’t know the solution, Then
numbers both can’t be prime simultaneously (but one of them still can)
Аs the sum guy knows that product guy can’t know the solution, then there sum must be odd
(if the sum will be even then numbers can be prime and there is an opportunity that P guy knows the answer)
If the sum is odd then one of the numbers must be odd (let it be a) and the other is even (call it b)
Then the product will be a*b or 2*a*(b/2)
The P guy must know the separate numbers. But there can be two pairs from the last equation.
He can choose pair 2*a and b/2, or a and b. And there is only one point when these pairs will be the same thing – when a=b/2. Then b=2a and their sum will be
a+b=3a
a
And we already know, that it can’t be even (because 2a is even already)
So there is an odd row from 3 to 31
3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
a can’t be prime Because if it will be than P guy will know the solution without asking S guy
so it leaves
9 15 21 25 27 for a
And
18 30 42 50 54 for b
And only the pair 25 and 50 has the product (1250=2*5*5*5*5) that can be factorize in only one way (2*5*5) and 5*5 that fit for all conditions
Am I wrong?