If a prime and its square both divide a number n, prove that $n=a^2 b^3$ Lets call a number $n$ a fortified number if $n>0$ and for every prime number $p$, if $p|n$ then $p^2|n$. Given a fortified number, prove that there exists $a,b$ such that $n=a^2b^3$.
I know that this must revolve around the fundamental theorem of arithmetic, but I can't figure out how to do this proof.
Some examples of fortified numbers: 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200
 A: You're correct this is related to the fundamental theorem of arithmetic. The statement regarding a powerful or squarefull (as Gerry Myerson's question comment suggests naming them instead of fortified) number means all of its prime factors must have a power of $2$ or larger.
You're asking to prove that all such numbers $n$ can be expressed using natural numbers $a$ and $b$ where
$$n = a^2b^3 \tag{1}\label{eq1A}$$
This involves basically proving that for every integer $m \ge 2$, there are non-negative integers $c$ and $d$ such that
$$m = 2c + 3d \tag{2}\label{eq2A}$$
You state in your comment you can do this, so I'll leave that to you, as it's fairly easy to do, such as with induction. Also note it's related to a particular case of the coin problem which shows the maximum number that can't be expressed by \eqref{eq2A} is $a_1 a_2 - a_1 - a_2 = 2(3) - 2 - 3 = 1$, so anything $2$ or larger can.
With this, consider each prime factor $p$ of $n$. Consider the p-adic valuation function $v_p(m) = e$ which gives, for any prime $p$, the highest exponent $e$ such that $p^{e} \mid m$. Then for any prime factor $p$ of $n$, have $v_p(n) = m$, with $v_p(a) = c$ and $v_p(b) = d$. Do this for all $p \mid n$ to create the corresponding prime factors of $a$ and $b$, and with both $a$ and $b$ not having any other prime factors. You will then end up with \eqref{eq1A}.
A: Hint $\ n = a^2 b^3\, $ for $\,\color{#c00}b=$ product of primes that occur to $\rm\color{#c00}{odd}$ power in $\,n\,$ and $\,a= \sqrt{n/b^3},\,$ an integer since all primes in $\,n/b^3\,$ occur to even power, e.g. $\,\color{#c00}2^{\large 7} 3^{\large 4}\color{#c00}5^{\large 3} 7^{\large 2}\! = (2^{\large 2} 3^{\large 2} 7)^{\large 2}\,(\color{#c00}{2\cdot 5})^{\large 3}$ 
