Show that $n$ is square-full iff $n$ can be written in the form :
$$ n = a^2 b^3 $$
I stumbled upon this as an exercise, but I do not see how I should attempt to show it. If we have $b=1$ , shouldn't this be obvious ? Should I try to do it for different values of $b$ ? I'd appreciate a hint more than a full answer, since I would like to do it myself. Also, how would one generalize this to numbers that are k-full? Thanks in advance.
On
Let r be the power of p dividing n. Consider the cases where r is even and odd. The hint is that by a clever labelling of r, you can easily see that: you get a product of primes with even power in the first case and a product of cubes in the second case, hence the desired result.
On
Suppose, a number $n$ is squareful. The prime factorization of $n$ is $$p_1^{a_1}\cdots p_n^{a_n}$$ with $a_1,\cdots a_n\ge 2$
Now, every number $k\ge 2$ can be expressed in the form $2u+3v$ with non-negative integers $u$ and $v$, therefore we can seperate every prime power into squares and cubes.
Assume that $n=a^2b^3$. It is obvious that a prime factor of $a$ or $b$ must have at least exponent $2$ in the prime factorization.
On
$n=1.$ Trivial. $(a=b=1.)$
$n>1.$ Let $n=\prod_{p\in S}p^{e(p)}$ where $S$ is the set of prime divisors of $n.$ Each $e(p)\geq 2.$
If $e(p)$ is even let $f(p)=e(p)$ and $g(p)=0 $. If $e(p)$ is odd let $f(p)=e(p)-3$ and $g(p)=3$. Then $$n=\left(\prod_{p\in S}p^{f(p)/2}\right)^2\cdot \left(\prod_{p\in S}p^{g(p)/3}\right)^3.$$
Think about the prime factorization of $n$. If $p^k$ is the highest power of $p$ that divides $n$, then $k\geq 2$ since $n$ is square-full. If $k$ is even, then $p^k$ is a square. If $k$ is odd, then it is at least 3, so it's of the form $2m+3$, where $m\geq 0$. Then $p^k = p^{2m}p^3$ which is a square times a cube. So you can express the prime factorization of $n$ as a square times a cube.