4A. The wise man $S$ was only told the sum of three consecutive natural numbers, and the wise man $P$ was only told their product, but neither of them was able to determine the original three numbers.

“If I knew,” said $S$, “that your number is greater than mine, I would determine the numbers.”

“But my number is smaller than yours,” concluded $P$, and “the numbers are ..., ... and ...”.

Determine the three numbers.
Solution. Let's denote the sum of the three numbers by $s$, and the product by $p$. It is clear that $s \geq 3$.

Sage $S$ knows the sum $s$, but it is such that he cannot determine the numbers knowing only this sum without additional information.

If $s=3$, the three given numbers are all 1, in which case Sage $S$ would immediately determine the numbers. Therefore, $s$ is not 3.

If $s=4$, the three numbers are 2, 1, and 1, but in this case, Sage $S$ would also immediately determine the numbers. Therefore, $s$ is not 4.

If $s=5$, the three numbers are 3, 1, and 1, or 2, 2, and 1. In the first case, $p=3$, and in the second case, $p=4$. In both cases, $p < s$:

in the first:

$$
p=3 < 5=s
$$

and in the second:

$$
p=4 < 5=s
$$

In both cases, the product is less than the sum, so Sage $S$ could not determine the numbers even if he knew that the product is less than the sum (there are at least two possibilities). Therefore, $s \neq 5$.

If $s=7$, the three numbers are 5, 1, and 1, or 4, 2, and 1, or 3, 3, and 1, or 3, 2, and 2. In the first case, $p=5$, in the second $p=8$, in the third $p=9$, and in the fourth $p=12$. In the first case, $p < s$:

$$
p=5 < 7=s
$$

and in the second:

$$
p=2 s-16=s+s-16 \geq s+7-6=s+1>s.
$$

In both cases, the product is greater than the sum, so Sage $S$ could not determine the numbers even if he knew that the product is greater than the sum (there are at least two possibilities). Therefore, $s \leq 6$.

The only remaining possibility is $s=6$. In this case, the three numbers are 4, 1, and 1, or 3, 2, and 1, or 2, 2, and 2. In the first case, $p=4$, in the second $p=6$, and in the third $p=8$. Sage $S$ knows these three possibilities and, if he knows that the product is greater, he can conclude that the three given numbers are 2, 2, and 2. Therefore, $s=6$.

Sage $P$ thinks in the manner described above and concludes that $s=6$. $P$ knows the product and states that it is less than the sum. Therefore, $p=4$ and the given numbers are 4, 1, and 1.