Adel Mardelli

as far as I can tell, we can only hope to get families of pairs, and not a definite answer. here's the best that I could come up with: any pair of numbers of the form: 2^n, prime number will do. try it?! if S knows that P cannot have the answer, then P must be looking at an odd sum, since if the sum was en even number, it could be broken down into a sum of two primes (among other pairs). In that case, S cannot know FOR SURE that P will not be looking at a product that only breaks down into two primes. Since S is sure, the P must know that the sum S is an odd number. With that in mind, we now look at the product. Any number, the product P in particular, should be broken down into its prime divisors. In general, this can be expressed as: P = 2 x 2 x 2 x .... x p1 x p2 x p3 x... where it breaks down into a certain number of 2's and a certain number of odd primes. It can be easily checked that only those combinations with one odd prime divisor will yield to number that add to an odd sum, which S must have seen in order to ascertain that P cannot possibly hope to find the numbers (our starting statement). We can also rule out the Products P that can be divided by a single 2, since in that case 2xp can be immediately recognized by Mr. P. As a generalization, any pair of numbers of the form 2^n and prime number will satisfy the conversation. I hope this is anywhere near accuracy. Adel