Why must a powerful number be a product of a square and a cube?

Question

Given the definition of a powerful number (a number where, for every prime factor p, p^2 is also a factor), I would think all powerful numbers would be of the form m = (ab)^2 where a and b are integers and at least one is a prime number. However, every definition I have seen for powerful numbers states the form as m = (a^2)(b^3). Why?

Answer 1

The Mathworld Wolfram page says the same thing, that powerful numbers are always of the form a^2 * b^3, then it gives a link to A001694 in the On-line Encyclopedia of Integer Sequences which lists 36, 100, 196, 484, ... as powerful numbers. These are only of the form a^2 * b^3 if b = 1, which seems to take away much of the usefulness of the special form quoted.

Answer 2

8 is a powerful number (There is only one prime factor, 2, and 2^2 is also a factor.) But 8 is not a perfect square.
In this case, a=1 (not a prime, but an integer) and b=2 .

Answer 3

"A powerful number is a product of a square and a cube, which concurrently relate "a linear (length-->x) digital number application", or "2 Dimensional (area-->yx) digital number application or "3Dimensional (volume-->zyx) digital number application". Said versatilile state is a virtue of all numbers that are ( product of "a square and a cube")!

With regards!