$\gcd(a,b)=1, x^a = y^b\Rightarrow x = n^b$, $ y = n^a$ for an integer $n$.

Question

If $ a$, $ b$, $ x$, $ y$ are integers greater than $1$ such that $ a$ and $ b$ have no common factor except $1$ and $ x^a = y^b$ show that $ x = n^b$, $ y = n^a$ for some integer $ n$ greater than $1$.

I started this way: $x^a=y^b$ $=>$ $x=y^{b/a}=(y^{1/a})^b$. Thus $x$ is an integer, and so is $y^{1/a}$, so $y=n^a$ for some integer $n$ greater than $1$. Then from the given relation, we get $x=n^b$.

I am not very convinced with my solution, as I feel I did some mistake. Can anyone point out where the mistake is? Any other solution is also welcome :)

Answer 1

Here is an alternate method.

Note that set of primes dividing $x$ and $y$ are same. Take any prime $p$ diving $x$(and hence $y$), and let $\alpha$ is the maximum power of $p$ in $x$ and $\beta$ is the maximum power of $p$ in $y$. Then $x^a=y^b \implies p^{\alpha a}=p^{\beta b}$, which implies $a|\beta b $ and $b| \alpha a$. But remember that $gcd(a,b)=1$. So, $a|\beta$ and $b|\alpha$.

Suppose, $\beta= a\cdot \beta_p$ and $\alpha=b\cdot \alpha_p$. Then we have, $\alpha a=\beta b$ or, $b\alpha_pa =a\beta_p b$ or, $\alpha_p=\beta_p$. So, for each prime $p$ diving $x$, we have such $\alpha_p$. Check that $n=\prod_{p|n}p^{\alpha_p}$ satisfies the required property.

Answer 2

Hint You can find some $k,l \in \mathbb Z$ such that $$ka+lb=1$$

Then $$n=n^{ka+lb}=(n^a)^k\cdot(n^b)^l$$

Now, if you want $x = n^b, y=n^a$ then the only possible choice for $n$ is $$n=??$$

Just prove that this is an integer and that this choice works.

Answer 3

Mimic $\rm\color{#c00}{subtractive}$ Euclidean algorithm on $\color{#c00}{(a,b)}.\,$ Clear if $\,a\!=\!b\,$ by $\,a,b\,$ coprime $\Rightarrow\, a\!=\!b\!=\!1$

Else wlog $\,a > b\,$ therefore $\ x^{\large\color{#c00}{a-b}} = (y/x)^{\large \color{#c00}b}\,$ so $\,y/x\in\Bbb Z\,$ by the Rational Root Test.

By induction we conclude: $\, x = n^{\large b},\,\ y/x = n^{\large a-b} \Rightarrow\, y = n^{\large a}\ \ \ {\bf\small QED}$

Answer 4

This is a $\rm\color{#0a0}{multiplicative}$ form of the following well-known $\rm\color{#90f}{additive}$ result about reduced fractions. A proof follows immediately by translating from additive to multiplicative form $\,n\cdot x\to x^n,\,$ i.e.

If $\,\ \overbrace{\gcd(a,b)=1}^{\textstyle aj+bk\,=\,1}\ $ then $\,\ \overset{\rm\large Unique\ Fractionization_{\phantom{|}}\!}{\bbox[5px,border:1px solid red]{\dfrac{y}x = \dfrac{a}b\:\Rightarrow\begin{align}\,y = na\\ x = nb\end{align}}}\ \ \ {\rm for\ some}\,\ n\in\Bbb Z, \, $ with proof as below

$\ \ \begin{align} \color{#c00}{xa=yb}\,\Rightarrow\,x &= \color{#c00}x(\color{#c00}aj\!+\!bk) = (\color{#c00}yj\!+\!xk)\color{#c00}b = n b\ \ \ \ \color{#90f}{\text{[additive]}}\\[.3em] \color{#c00}{x^{\Large a}= y^{\Large b}}\Rightarrow\,x &= \color{#c00}x^{\Large \color{#c00}aj\,+\,kb}\ =\ (\color{#c00}y^{\Large j}\! \cdot x^{\Large k})^{\Large \color{#c00}b} = n^{\Large b}\ \ \ \color{#0a0}{\text{[multiplicative]}} \end{align}$

Note $\, n = y^{\large j} x^{\large k}\,$ is a rational root of $\,n^{\large b} = x\in\Bbb Z,\,$ so $\,n\in\Bbb Z\,$ by the Rational Root Test.

Remark $ $ The analogy between additive and multiplicative forms is clarified when we study abelian groups as $\,\Bbb Z\!\!-\!\!\text{modules}$. Said fundamental result about principality of fractions is sometimes called Unique Fractionization to emphasize its equivalence with uniqueness of prime factorizations.