Which two-digit (integer) number is equal to the product of the sum and difference of its digits?
I. solution: Let the sought positive number be $\overline{x y}=10 x+y$, that is, $1 \leq x \leq 9$ and $0 \leq y \leq 9$. The problem does not specify which digit is the minuend in the difference of the digits, so we must consider two possibilities:

a) either $10 x+y=(x+y)(x-y)=x^{2}-y^{2}$,

b) or $10 x+y=(x+y)(y-x)=y^{2}-x^{2}$.

Case a) is not possible, because it leads to $y^{2}+y=y(y+1)=x^{2}-10 x=x(x-10)$, where the left side is non-negative, while the right side is negative. Therefore, $y>x$, so $y \geq 2$.

From case b), with similar rearrangement,

$$
x(x+10)=y(y-1)
$$

and here the right side is even, as the product of two consecutive integers, so both factors on the left side must be even, because $x$ and $x+10$ are either both even or both odd; thus, the left side is also divisible by $2^{2}=4$. In fact, it is divisible by $2^{3}=8$, because either $x$ is divisible by 4, or $x+10$ is. The right side, however, is only divisible by 8 if either $y=8$, or $y-1=8$.

From (1), we only get a positive integer solution for $x$ with $y=8$: $x=4$, so the sought number is 48. - Indeed, $48=(4+8)(8-4)$.

Rozália Fekete (Celldömölk, Berzsenyi D. g. II. o. t.)

II. solution: The case a) is not possible because $10 x+y \geq 10 x>x^{2} \geq x^{2}-y^{2}$. We rewrite the equality under case b) as follows:

$$
11 x=(x+y)(y-x)+x-y=(x+y-1)(y-x)
$$

From this,

$$
x=\frac{(x+y-1)(y-x)}{11}
$$

is an integer, and since $1 \leq y-x \leq 8$, and every integer in this range is relatively prime to 11, it follows that $x+y-1$ is divisible by 11. Since $y \leq 9, x<y$, that is, $x \leq 8$, we have $x+y-1 \leq 16$, and from $x \geq 1, y \geq 2$ we have $x+y-1 \geq 3$, so the only possible value for $x+y-1$ is 11. Now we have $x+y=12$ and from (2) $x=y-x, y=2 x$, so $x=4, y=8$, thus the sought number is 48.