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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 :)

Bill Dubuque
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Yellow
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4 Answers4

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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.

N. S.
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  • $n=y^kx^l?$ Is that what you mean? – Yellow Jan 08 '19 at 04:40
  • But $k$ and $l$ may be negative right? – Yellow Jan 08 '19 at 17:51
  • @AnuRadha True. But with this choice you can show that $x=n^b, y=n^a$ and deduce from here that $n$ is integer. – N. S. Jan 08 '19 at 18:18
  • The precise intent of the hint will likely be far from clear for many readers. Please provide further details. – Bill Dubuque Jan 09 '19 at 16:24
  • @AnuRadha I added an answer which shows how to derive a Bezout-based proof immediately by translating it from its well-known additive form (a well-know result about fractions). This is probably closely related to what is intended above, but I can't be sure since the author has not provided further details. – Bill Dubuque Jan 10 '19 at 18:35
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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.

tarit goswami
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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 $\rm\color{darkorange}{translating}$ from additive to multiplicative form $\,n\cdot x\to x^n,\,$ i.e.

If $\,\ \overbrace{\gcd(a,b)=1}^{\textstyle aj+bk\,=\,1}\ $ then $\,\ \overset{\bf\large Unique\ Fractionization_{\phantom{|}}\!}{\bbox[5px,border:1px solid red]{\dfrac{y}x = \dfrac{a}b\:\Rightarrow\begin{align}\,y = na\\ x = nb\end{align}}}\ \ \ {\rm for\ some}\,\ \color{#0af}{n\in \Bbb Z}\ (\rm vs. \Bbb Q)$

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

$ n = y^{\large j} x^{\large k}\!\in\Bbb Q\,$ is a rational root of $\,n^{\large b} = x\in\Bbb Z,\,$ so $\,\color{#0af}{n\in\Bbb Z}\,$ by Rational Root Test.

Generally, ignoring final line $\rm\color{#0af}{using\ RRT}$, the proof shows that in any abelian (commutative) group we have: $\ z = x^a = y^b \,\Rightarrow\, x = n^b,\,$ for $\, n = y^jx^k,\,$ so $\,z = x^a = n^{ab},\,$ i.e.

Theorem $ $ If $\:\!z\:\!$ is both an $\:\!\color{#c00}a$'th & $\:\!\color{#c00}b$'th power then $\:\!z\:\!$ is an $\color{#c00}{ab}$'th power, if $\,a,b\,$ are coprime.
[note $\color{darkorange}{\rm C}$ = commutativity is used: $\,x^{aj+kb} = (\color{#c00}{x^a})^j x^{kb}\! = (\color{#c00}{y^b})^j x^{kb} \overset{\color{darkorange}{\rm C^{\vphantom |}}}= (y^j x^k)^b\! = n^n$].

The above analogy between additive and multiplicative forms will be clarified when one studies group theory and $\:\!\Bbb Z\!\!-\!\!\text{modules}$.

Said fundamental result about principality of fractions is sometimes called Unique Fractionization to emphasize its equivalence with uniqueness of prime factorizations.

Bill Dubuque
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  • Or replacing use of Bezout by the gcd distributive law $$,x(a,b) = (\color{#c00}{xa},xb) = (\color{#c00}{yb},xb) = (y,x)b =: nb\qquad$$

    i.e. $q = \frac{y}a = \frac{x}b,$ is writable two ways with coprime denom's therefore $,q,$ is an integer $,n.\ $ $\bf Or$ we can cancel $,n=(x,y),$ to reduce $,y/x,$ then infer $,y/n = a,\ x/n = b,$ by unique fractionization (wlog $,x,b>0)\ \ $

     – Bill Dubuque Sep 04 '25 at 13:49
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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: $\, x = n^{\large b},\,\ y/x = n^{\large a-b} \Rightarrow\, y = n^{\large a}\ \ \ {\bf\small QED}$

Bill Dubuque
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  • Induction? What do you induct on? – Yellow Jan 09 '19 at 16:51
  • @Anu Same as in the subtractive Euclidean algorithm, e.g as there we can use the "height" $,h(a,b) := a+b,$ Any $h$ that descreases when we subtract its smaller argument from the larger also yields descent (e.g. we can also choose $,h =$ max, product, etc) – Bill Dubuque Jan 09 '19 at 17:44
  • Your answer is nice, though it takes some time for me to realize what is actually going on. (Umm, I am new to such kind of inductive stuff). Anyways, thanks. I accept your solution. – Yellow Jan 09 '19 at 17:58