Show that if a and b are positive integers, and $a^3 | b^2$, then a | b Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.
If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3r ≤ 2s. And so we have, r ≤ (2/3)s < s.
But how come this implies that $p^r | b$, and eventually a | b.
 A: Let $a,b$ be elements of $Z^+$  (i.e., the positive integers) and suppose $a^3|b^2$.
Then, by definition $b^2=c*a^3$ for some $c$ such that $c$ is an element of $Z^+$.
Hence, $\sqrt{b^2} = b = \sqrt{c*a^3} = (\sqrt{a^2})*(\sqrt{c*a}) = a*\sqrt{c*a}$
This works for both roots of $b^2$ and $a^2$.  
Consider case-by-case:
$\sqrt{b^2}$ = $b$ & $\sqrt{a^2}$ = $a$:   $b = a*(\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2}$ = $a$:  -$b$ = $a*(\sqrt{c*a}) \Rightarrow  b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ =$ b$ & $\sqrt{a^2} = -a:  b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2} = -a: -b = -a*(\sqrt{c*a}) \Rightarrow b = a*(\sqrt{c*a})$
Hence, $a|b$ by definition.
A: Let 
$$a = p_1^{q_1}p_2^{q_2}\dots p_n^{q_n}$$
where each $p_i$ is a prime and each $q_i$ is a non-negative integer
Similarly, let
$$b = p_1^{q'_1}p_2^{q'_2}\dots p_n^{q'_n}$$
where each $q_i$ is a non-negative integer
Since $a^3|b^2$, 
$$\implies p_1^{3q_1}p_2^{3q_2}\dots p_n^{3q_n}|p_1^{2q'_1}p_2^{2q'_2}\dots p_n^{2q'_n}$$
This means, for each $p_i$, 
$$3q_i \leq 2q'_i$$
Now you just need to prove, for each $p_i$,
$$q_i \leq q'_i$$
which is obvious:
$$3q_i \leq 2q'_i$$
$$\implies q_i \leq \frac23\cdot q'_i \leq q'_i$$
A: Since $a^2\mid a^3$, we have $a^2\mid b^2$. Then let $c=b/a$. We want to show $c\in \mathbb{Z}$. Note that $c$ satisfies $x^2-\frac{b^2}{a^2}$. This is monic in $\mathbb{Z}[x]$, and since any rational solution to a monic in $\mathbb{Z}[x]$ is actually an integer, we get that $c\in\mathbb{Z}$. 
Thm: If $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ is such that $a_i\in \mathbb{Z}$, and we have that for some $c\in \mathbb{Q}$ $f(c)=0$, then $c\in \mathbb{Z}$
Proof: Write $c=b/a$ with $c$ a reduced fraction (meaning $a$ and $b$ are coprime). Then we have: 
$$ 
(b/a)^n+a_{n-1}(b/a)^{n-1}+...+a_0=0
$$Multiply everything by $a^n$. You get that $b^n=aD$ for $D\in \mathbb{Z}$. Hence, $a\mid b^n$. If $a$ is not $\pm 1$, then there is a prime $p$ such that $p\mid a$. This means that $p\mid b^n$ which then implies that $p\mid b$. This contradicts the fact that $a$ and $b$ are coprime. Hence we get that $a=\pm 1$, so $c\in \mathbb{Z}$ just as we wanted. 
A: We can prove by more basic postulate.
if we defined gcd of a & b as the minimum of $ax+by$,
let $g = gcd(a,b)$, then we can find $x \& y$ such that $ax+by = g$ and of coarse $g|a*1+b*0,a*0+b*1$ because the set of $ax+by$ is the set of $g,2g,...$.
let $a=ga_1, b=gb_1$,$b^2=a^3k$ then $ax+by=g$ comes to $a_1b_1x+ga_1^3ky=b_1$
so $a_1|b_1$ and $ga_1|gb_1$
