following my logic, one has to be odd and one has to be even... that is the only way S can know anything about the two numbers. because the sum is odd, S knows that at least one of the numbers is not prime, and therefore P cannot know the answer.
now, assuming that is correct, start writing a table of sums with the even integers in a column and the odd numbers in a row.
obviously, the numbers cannot be (3,4), (4,5), (3,6), (4,7), (5,6) because the product would be 12, 20, 18, 28, 30 and P would know the answer.
the first pair that would not be obvious to P and can still be determined by S are (8,3). the prodcut for P is 24 and there are only 2 possiblilities (8,3) and (4,6).
the last sum that can be determined by S is 11 because there are only 3 pairs (3,8), (4,7) and (5,6) that give the sum... after that, there are too many combinations and S would never know the answer.
so, given (3,8), we have
product = 24, combinations (3,8) (4,6)
sum = 11, combinations (3,8) (4,7) (5,6)
when P says, "i do not know the numbers," S knows that the pair is not (7,4) because that would be 28 and that is the only combination for the product. which leaves (3,8) and (5,6) for S.
when S says, "i know, you won't know the numbers," P knows that at least one number is odd, so that only leave (3,8) for P.
when P says he knows the numbers, S realizes that the numbers are (3,8) because (5,6) would give 30 and P would not have been able to chose between (5,6) and (3,10).
if you write the first 6 or 7 columns and rows of the addition table, you will quickly see that this is the only way S could ever chose 1 pair of numbers. there are numbers at the high end of the table, but those pairs generate unique products and P would have known the answer immediately.
i hope this is clear as mud (and that i am correct)
[ September 03, 2003: Message edited by: Greg Harris ]