8.79 Given that the numbers 1 and 2 are written on the blackboard. Now, it is allowed to write a new number according to the following rule: when the numbers $a$ and $b$ are on the blackboard, the number $a b + a + b$ can be written. Can the numbers 13121 and 12131 be written on the blackboard?
[Solution] Let $c=ab+a+b$, then $c+1=(a+1)(b+1)$. This means that if we add 1 to all the numbers that appear at the beginning and during the operation, the problem transforms into starting from 2 and 3, and each new number written is the product of the original two numbers. Therefore, all the numbers that appear on the blackboard are of the form $2^{n} \cdot 3^{m}$, where $m$ and $n$ are non-negative integers, and except for the original two numbers, $m$ and $n$ are natural numbers. Conversely, any number of the form $2^{n} \cdot 3^{m}$ can appear on the blackboard. Thus, under the original operation, the numbers obtained on the blackboard are all of the form $2^{n} \cdot 3^{m}-1$. Noting that $13121=2 \cdot 3^{8}-1, 12131=2^{2} \cdot 3^{2} \cdot 337-1$, we see that 13121 can be written on the blackboard, but 12131 cannot.