Sum and product puzzle

Mr. Puzzle: I choose two natural numbers greater than 1. I will tell the sum of the numbers only to Mr. Sum, and their product only to Mr. Product.
He tells them.
Mr. Product: I do not know the numbers.
Mr. Sum: I knew you didn’t.
Mr. Product: But now I know!
Mr. Sum: So do I!

What can be the numbers if they are not greater than 100?

4 and 13.

Why (short version):

• Mr. Product doesn’t know ⇒ the product has more than one factorization in 2..100 (so not a prime, not a prime square like $49=7\cdot7$, etc.).
• Mr. Sum “knew” that ⇒ for every split of his sum $S$ into two numbers $\ge 2$, the product would still be ambiguous. That forces $S$ to be one of the sums that cannot be written as prime + prime (within 2..100). Those are $\{11,17,23,27,29,35,37,41,47,53\}$.
• Mr. Product now knows ⇒ among his product’s factor pairs, exactly one has a sum from that set. For $52$: $52=2\cdot26$ (sum 28, not in the set) or $52=4\cdot13$ (sum 17, in the set). So Product deduces $4$ and $13$.
• Mr. Sum now knows ⇒ with $S=17$, the splits are $(2,15),(3,14),(4,13),(5,12),(6,11),(7,10),(8,9)$. Only $(4,13)$ has a product (52) that would let Product deduce uniquely at the previous step; the others (e.g., $30=2\cdot15=3\cdot10=5\cdot6$) still leave Product with multiple “good” sums.

Hence the unique pair ≤100 is (4, 13).