Exercise 1.3.15 (Conditions for $a$ to be a cube? For $a^2 \mid b^2$?)

Proof.

(a)

As in (1.6), $$a=\underset{p\in A}{\prod }{p}^{\alpha (p)},b=\underset{p\in A}{\prod }{p}^{\beta (p)}.$$

(We can take the same set $A$, if we put $\alpha (p)\ge 0,\beta (p)\ge 0$.)

Suppose that $a=c^3$ is a cube, where $c=\underset{p\in A}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{\gamma (p)}$ (Every $p$ such that $p\mid c$ satisfies $p\mid A$, so we can take the same set $A$). Then

$$a=\underset{p\in A}{\prod }{p}^{\alpha (p)}=\underset{p\in A}{\prod }{p}^{3\gamma (p)}.$$

The unicity of the factorization in prime factors shows that $\alpha (p)=3\gamma (p)$ for every $p\in A$. Thus

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in A,3\mid \alpha (p).$$

Conversely, suppose that $3\mid \alpha (p)$ for every $p\in A$. Then $\alpha (p)=3k_p$, where $k_p\in \mathbb{N}$, thus

$$a=\underset{p\in A}{\prod }{p}^{\alpha (p)}=a=\underset{p\in A}{\prod }{p}^{3k_p}=a={\left(\underset{p\in A}{\prod }{p}^{k_p}\right)}^3,$$

so $a$ is a cube.

To conclude, if $a=\underset{p\in A}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{\alpha (p)}$,

$$\exists <!--\; FUNCTION\; APPLICATION\; -->c\in \mathbb{N},a=c^3⟺\forall <!--\; FUNCTION\; APPLICATION\; -->p\in A,3\mid \alpha (p).$$

(b)

If $a^2\mid b^2$, then there is some integer $q$ such that $b^2=q{a}^2$. Every prime factor $p$ of $q$ divides $b$, so is an element of $A$. Write $q=\underset{p\in A}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{\delta (p)}$. Then $b^2=q{a}^2$ gives $$\underset{p\in A}{\prod }{p}^{2\beta (p)}=\underset{p\in A}{\prod }{p}^{\delta (p)+2\alpha (p)}.$$

The unicity of the factorization in prime factors shows that, fro every $p\in A$,

$$2\beta (p)=\delta (p)+2\alpha (p).$$

Therefore $\delta (p)$ is even, so $\delta (p)=2𝜀(p)$ for some integer $𝜀(p)$. This gives

$${\left(\underset{p\in A}{\prod }{p}^{\beta (p)}\right)}^2={\left(\underset{p\in A}{\prod }{p}^{𝜀(p)+\alpha (p)}\right)}^2,$$

where the products are positive, so that

$$\underset{p\in A}{\prod }{p}^{\beta (p)}=\underset{p\in A}{\prod }{p}^{𝜀(p)}\underset{p\in A}{\prod }{p}^{\alpha (p)}.$$

This shows that $b=\mathit{ka}$, where $k={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{p\in A}{p}^{𝜀(p)}$ is an integer, thus $a\mid b$.

Conversely, if $a\mid b$, then $b=\mathit{ka}$ for some integer $k$, thus $b^2=k^2{a}^2$, so $a^2\mid b^2$. We have proved, for all positive integers $a,b$,

$$a^2\mid b^2⟺a\mid b.$$

The conditions satisfied by the exponents are

$$a^2\mid b^2⟺a\mid b⟺\forall <!--\; FUNCTION\; APPLICATION\; -->p\in A,\alpha (p)\le \beta (p).$$