Once there were 3 friends (Anton, Simon, Paul) chatting to each other in a dining hall. They were talking about numbers:

    A: I've chosen two natural numbers which are both greater than 1. I'm going to tell Simon the sum and Paul the product of these numbers.

Anton then gave Simon the sum and Paul the product, such that each of our friends knew "his" number only but not the number of the other friend.

    A: Now, can you guess the numbers which I have in mind?

Simon and Paul liked this sort of game. Their conversation was the following:

I. Simplified version:

    P: I don't know Anton's numbers.
    S: Well, now I do!
    P: Then I know them too.

A math professor had observed this whole situation from the very beginning.  He went to our friends and said "I know Anton's numbers, too!"  Do you?

Answer to simplified version:

Because Paul can not deduce the two natural numbers (n,m, each greater than 1) from their product P, therefore everyone knows that the smallest possible value  of P is 12.  The next possible values are 16, 18, 20, 24, ... (numbers which can be written as products of two factors in more than one way.)  Simon now knows the answer.  Therefore the sum S has to be such that there is only one way to find two numbers n,m, with n+m=S and n x m = P = 12 or 16 or ... .  S=7=3+4 is the only possibility.  If S were any bigger, there would not be a unique choice of n and m.  So Anton's numbers are 3 and 4.

II. Full version

    P: I don't know Anton's numbers.
    S: I don't know them either.
    P: Hey, now I know.
    S: Then I know them too.

A math professor had observed this whole situation from the very beginning.  He went to our friends and said "I know Anton's numbers, too!"  Do you?

Answer to full version:

This is different!  Simon is not able to find a unique pair that sums to S and multiplies to 12 or 16 or ... .  Therefore the pair (3,4) with S=7 can't be right.  This is the extra clue which enables Paul to find the answer.  If knowing that (3,4) is wrong tells him what the right answer is, then P must be 12 and the pair must be (2,6).  Any other value of P would not allow Paul to make much use of Simon's inability to solve the problem.  Note that S=8 does not enable Simon to decide between (2,6) and (4,4).

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P. B. Allen 5/2/2003