Exercise 1.3.19 (Perfect squares)

Question

Let $a$ and $b$ be positive integers such that $(a,b) = 1$ and $ab$ is a perfect square. Prove that $a$ and $b$ are perfect squares. Prove that the result generalizes to $k$th powers.

Richard Ganaye · Accepted Answer

\begin{proof} Write
$$a = \prod_{p \in A} p^{\alpha(p)},\qquad b = \prod_{p \in B}p^{\beta(p)},$$
where $\alpha(p)>0, \beta(p)>0$.

Since $a \wedge b = 1$, $A \cap B = \varnothing$ (otherwise, if $p \in A\cap B$, $p \mid a$ and $p \mid b$, thus $a \wedge b >1$). Define $\gamma(p)$ by  
$$
\left\{
\begin{array}{ll}
\gamma(p) = \alpha(p)& 	ext{if } a \in B,\
\gamma(p) = \beta(p)& 	ext{if }a \in B.
\end{array}
ight.
$$
Then
$$ab = \prod_{p \in A \cup B} p^{\gamma(p)}.$$
By the text (p.25) or Problem 15, $ab$ is a perfect square if and only if $\gamma(p)$ is even for every $p \in A \cup B$. By definition of $\gamma(p)$, this shows that $\alpha(p)$ is even if $a \in A$, and $\beta(p)$ is even if $p \in B$. Therefore $a$ and $b$ are perfect squares.

\bigskip
More generally, if $ab$ is a $k$th power, then $k \mid \gamma(p)$ for all $p \in A \cup B$ (see Problem 15). Therefore $k \mid p$ for all $p \in A$, and $k \mid p$ for all $p \in B$, hence $a, b$ are $k$th powers.
\end{proof}

Exercise 1.3.19 (Perfect squares)

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Exercise 1.3.19 (Perfect squares)

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