Sum Things are Products of Twisted Minds

First Thoughts:

It may seem to the uninitiated observer that no concrete information was communicated between S and P in their discussion, but if we assume that both are completely rational agents, we can in fact figure out the numbers x and y. Each statement that S or P makes communicates vital information to the reader. We thus decide to proceed by examining them one at a time to decode their content. Before we even get this far, however, we may note that since x and y are both at least 2, we know that their sum and product are both at least 4. Furthermore, the product xy cannot be a prime number, since neither x nor y can be equal to 1.

 

Second Thoughts:

When S announces that he does not know what the numbers are, we know that the sum x + y must be realizable in at least two different ways as a sum of integers greater than or equal to 2. This eliminates x + y = 4 and x + y = 5, because each of these can only be expressed as such a some in a single way, up to order. These sums are 2 + 2 = 4 and 2 + 3 = 5 respectively. When we consider x + y = 6, we see that we could either have 4 + 2 = 6 or 3 + 3 = 6. Any larger sums will also be expressible in more than one way. Therefore, we have discovered that x + y is at least 6. This seems to be all we can glean from S at the moment.

 

Third Thoughts:

Next, we move on to P’s statement. He reveals two pieces of information: he knows that S did not know the two numbers, and he does not know the product. For the first of these, P is saying in effect that he knows that the sum x + y is at least 6, since he is able to follow the same chain of logic as S. Therefore, he knows that x and y are not equal to 2 and 2 or 2 and 3. It follows that our product is at least 6.

 

Examining the second statement, we see that if x and y are both prime numbers, then P would know the product. Therefore, it must be the case that at most one of x and y is prime. What does this mean for the product xy? Since at most one of x and y is prime, the product xy must have at least 3 prime factors, although these factors may not necessarily be unique. For example, we could have xy = 16 = 2 * 2 * 2 * 2 or xy = 12 = 2 * 3 * 3. Furthermore, we must also have that xy is not the cube of a prime. If xy = p^3, then the only possible decomposition, up to order, is x = p and y = p^2.

 

Final Thoughts:

We have come to the final exchange of information between the two interlocutors. S is able to ascertain from P’s clues the identity of the two numbers x and y. This might initially give us pause, because we are not able to make a similar discovery, but our situation is a result of us having neither knowledge of the sum nor product. S is more privileged than us, because he already knows the sum. Now we have to make sense of how S is able to figure out the numbers.

We already know that the sum S knows can be expressible in more than one way. If all of these decompositions of the sum resulted in both x and y being prime, then P would know the product. It follows that there is at least one decomposition where either x or y is not prime. If there was more than one decomposition where this was the case, then S would not be able to determine the numbers. Therefore, there is exactly one decomposition where x or y is not prime. All other decompositions must consist of prime numbers added together.

Let’s use this information to find an upper bound on our sum. The first two composite numbers larger than 2 are 4 and 6. Therefore, any integer that can be decomposed in two separate ways, one of which involves a 4 and one of which involves a 6, is too large to be our sum. Since x and y are both at least 2, we see that the smallest such integer is 2 + 6 = 8 = 4 + 4. Any larger integer n can be written as n = (n – 4) + 4 = (n – 6) + 6. Therefore, our sum is at most 7.

We only have two candidates for the sum to consider: 6 and 7. We see that there are two possible decompositions for each: 6 = 2 + 4 = 3 + 3 and 7 = 2 + 5 = 3 + 4. We consider them in this order. For the decompositions of 6, we see that 2 * 4 = 8 = 2^3 and 3 * 3 = 9. Neither of these products meet our established criteria for products, so we throw them out. This leaves 7. Since 2 and 5 are both prime, we must have x and y are 3 and 4, resulting in xy = 12.