Exercise 6.3.6 (When $x^n = a/b$ has a rational solution ?)

Question

Let $a/b$ be a positive rational number with $a>0,\ b>0$, $\mathrm{g.c.d}.(a,b) = 1$. Generalize Corollary 6.15 by proving that for any integer $n>1$ the equation $x^n = a/b$ has a rational solution if and only if both $a$ and $b$ are $n$th powers of integers.

\bigskip
Hint. If $(a/b)^{1/n}$ is rational, so is $b(a/b)^{1/n}$, which is a root of the equation $x^n = ab^{n-1}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
I don't use the hint.
\begin{itemize}
\item If $a = \alpha^n$ and $b = \beta^n$ are $n$-th powers of the integers $\alpha, \beta$, then $(\alpha/ \beta)^n = a/b$, so $x^n = a/b$ has a rational solution.

\item Conversely, suppose that $x^n = a/b$ has a rational solution $c/d$, where $c \wedge d = 1$, $d>0$. Then 
\begin{align}
bc^n = a d^n.
\end{align}
Therefore $d^n \mid bc^n$, and $d^n \wedge c^n = 1$, thus $d^n \mid b$. So there is some integer $\lambda$ such that $b = \lambda d^n$. Substituting in (1), we obtain $\lambda d^n c^n= a d^n$, where $d
e 0$, therefore
\begin{align*}
a &= \lambda c^n,\
b &= \lambda d^n,
\end{align*}
where $\lambda >0$ since $b>0, d>0$.

Since $\lambda \mid a, \lambda \mid b$, then $\lambda \mid a \wedge b = 1$, where $\lambda >0$, thus $\lambda = 1$, and
\begin{align*}
a &= c^n,\
b &= d^n,
\end{align*}
are $n$-th powers of integers.
\end{itemize}

\end{proof}

Exercise 6.3.6 (When $x^n = a/b$ has a rational solution ?)

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Exercise 6.3.6 (When $x^n = a/b$ has a rational solution ?)

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