Mfrhtyj

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 :)

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, $beta= acdot beta_p$ and $alpha=bcdot alpha_p$. Then we have, $alpha a=beta b$ or, $balpha_pa =abeta_p b$ 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.

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

Then
$$n=n^{ka+lb}=(n^a)^kcdot(n^b)^l$$

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.

Mimic $rmcolor{#c00}{subtractive}$ 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^{largecolor{#c00}{a-b}} = (y/x)^{large color{#c00}b},$ and $,y/xinBbb Z,$ via Rational Root Test.

By induction we infer $, x = n^{large b}, y/x = n^{large a-b} Rightarrow, y = n^{large a}$

This is a multiplicative form of the following well known result about fractions, and its proof follows immediately by translating well-known additive proofs into multiplicative form, for example:

$aj!+!bk=1, {xa=yb},Rightarrow, bbox[5px,border:1px solid red]{dfrac{y}x = dfrac{a}b,Rightarrow,begin{align},y = na\ x = nbend{align}} {rm for some}, ninBbb Z, $ with proof as follows

$begin{align}
color{#c00}{xa=yb},Rightarrow,x &= color{#c00}x(color{#c00}aj!+!bk) = (color{#c00}yj!+!xk)color{#c00}b = n b,, y = color{#c00}y(color{#c00}bk!+!aj) = (yj!+!color{#c00}xk)color{#c00}a = n a\[.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}, y = color{#c00}y^{Largecolor{#c00} bk,+,aj} = (y^{Large j}! cdot color{#c00}x^{Large k})^{Largecolor{#c00} a} = n^{Large a}
end{align}$

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

Remark $ $ The analogy between additive and multiplicative forms will be clarified when one studies abelian groups as $Bbb Z!-!$ modules. The fundamental result about principality of fractions is sometimes called Unique Fractionization to emphasize its close relationship with unique factorization.