Sorry, I can't understand the problem. According to me, any good number is perfect square and the problem is false then. Correct me, please.

Amir Hossein wrote: Sorry, I can't understand the problem. According to me, any good number is perfect square and the problem is false then. Correct me, please. If $p$ is a divisor of a good number, then $p^2$ is also divisor. $8$ is good because for all primes which divides $8$, their square will divide $8$. $9$ is another good number. So $8$ and $9$ form a pair of consequtive number.

The product of two good numbers is a good number. $8\times 9 = 72$ is a good number. $4\times 72 = 288$ is a good number. Perfect squares are good. $288+1 = 289 = 17^2$ is a good number. So let $a$ and $a+1$ be consequtive good numbers. $4a(a+1)$ is a good number. $4a(a+1)+1$ is a good number because $4a^2+4a+1 = (2a+1)^2$. So if there is one pair of consequtive good numbers, there are infinitely many of them. $8$ and $9$ are the first, so there are infinitely many of them.