A puzzle about a sum and product of two numbers

Question

The Gray Man wants to test The Hardy Boys.

He says to them, "I've selected 2 positive integers, both bigger than one." He then proceeds to reveal their total and product to Frank and Joe respectively, and asks, "Now, can you surmise the numbers that I've selected?". The Hardy Boys accept this challenge. Frank says, "I don't know Mr. Gray's numbers." Joe responds, "I don't know them either." Frank exclaims, "Hey! Now I know." Joe cries out, "Then I know them too." Fenton Hardy had been a silent observer of this scene. He goes to the brothers and declares, "I know Mr. Gray's numbers as well!"

What are The Gray Man's numbers?

Answer 1

Gray must have told Frank 6 or greater. (Or else Frank could conclude that the numbers are either $(2,2)$ or $(2,3)$ for 4, and 5 respectively.

Gray must have told Joe a number with 3 or more proper divisors. (Or else Joe can conclude by just finding out the two divisors.)

Suppose the true numbers are $(N,M)$.

Frank is left guessing at $$(2, N+M-2), (3, N+M-3), (4, N+M-4), ...$$ He knows Joe is told one of $$2(N+M-2), 3(N+M-3), ...$$

But given Joe's answer he can eliminate any of the choices which have too few factors. This must eliminate all but one choice. Then we cannot have had both of these guesses for Frank: $(4, N+M-4)$ and $(6, N+M-6)$, because neither of these can be eliminated by Frank with Joe's information. (Their products both have more than three factors due to the 4 and the 6).

Then $N+M-6 \lt 2 \implies N+M \lt 8$

We can see that $N+M = 6$ does not fit the story:
Then Frank guesses at $(2,4), (3,3)$
Through Joe, he can eliminate both of these options.

Try $N+M = 7$ Then Frank guesses at $(2,5), (3,4)$ After hearing Joe's information, he eliminates $(2,5)$ and knows $(3,4)$. He now knows Gray's numbers. Everyone else in the story then reads this web page and also knows the numbers. So this is the only case consistent with the story.

Answer 2

We're looking for two numbers, $x$ and $y$. Let's let $y$ be the bigger one (if they're not equal). We know $x$ and $y$ are integers, and they're both greater than $1$, so we have $y \geq x\geq 2$. Frank knows the sum $(x+y)$, and Joe knows the product $xy$.

The fact that Frank doesn't immediately know the answer indicates that $x+y \geq 6$, since a sum of $4$ or $5$ would uniquely determine $x$ and $y$, as $4 = 2+2$ and $5=2+3$. So the first few possibilities for the sum are $6, 7, 8, \ldots$.

The fact that Joe doesn't immediately know $x$ and $y$ indicates that $x$ and $y$ are not both primes, and also $xy$ is not the cube of a prime (since $xy=pq$ or $xy = p^2$ or $xy=p^3$ would uniquely determine $x$ and $y$ by unique factorization). So the first few possibilities for the product are $12, 16, 18, 20, 24, 28, \ldots$.

In particular, we now know that Frank was not given $6$ for the sum, since the possibilities $3+3$ and $2+4$ are both eliminated by their corresponding products: $3*3 = 9$ and $2*4 = 8$ are not allowed.

If Frank was given $7$ for the sum, he would now know that $x=3$ and $y=4$, since the possibility of $x=2$ and $y=5$ has now been eliminated.

If, on the other hand, Frank was given a sum of $8, 9, 10,$ or $11$, then Joe's information about the product would not be sufficient for Frank to determine the numbers: $$ 8 = 2+6 = 4+4 $$ $$ 9 = 3+6 = 4+5 $$ $$ 10 = 2+8 = 4+6 $$ $$ 11 = 2+9 = 3+8 = 4+7 = 5+6 $$ And looking further ahead:

Proposition: If $n \geq 12$, then there will always be multiple ways to write $n$ as a sum of two positive integers greater than $1$ where at least one is composite, and they are not both powers of the same prime.

Proof: If $n= 0 \mod 4$, so $n = 4b$, then $b \geq 3$. Then $(2b, 2b)$ and $(2b+2, 2b-2)$ are two pairs of (even) composite numbers which sum to $n$. Or if $n$ is even and $k=n/2\geq 6$ is odd, then $k\geq 7$. In that case, $(k+1,k-1)$ and $(k+3, k-3)$ are two pairs of (even) composite numbers whose sum is $n$.

On the other hand, if $n = 2k+1 $ is odd, then $k\geq 6$. Consider the pair $(k, k+1)$. The numbers $k$ and $k+1$ can't both be prime (since they have different parity and neither is $2$), and they are not powers of the same prime (since their difference is $1$). On the other hand, we can say the same about the pair $(k-2, k+3)$: As above, they are not both prime. Further, they are not powers of the same prime since their difference is divisible by $5$ and therefore the prime would have to be $5$, but the difference between distinct positive powers of $5$ is always at least $20$.

So it seems to me that when Frank announces he has the answer, that by itself should be enough for us (and Fenton Hardy) to deduce the solution; further testimony from Joe is not required.