# If $a^m=b^m$ and $a^n=b^n$ for $(m,n)=1$, does $a=b$? [duplicate]

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Prove that $a=b$, where $a$ and $b$ are elements of the integral domain $D$

Something I'm curious about, suppose $a,b$ are elements of an integral domain, such that $a^m=b^m$ and $a^n=b^n$ for $m$ and $n$ coprime positive integers. Does this imply $a=b$?

Since $m,n$ are coprime, I know there exist integers $r$ and $s$ such that $rm+sn=1$. Then $$a=a^{rm+sn}=a^{rm}a^{sn}=b^{rm}b^{sn}=b^{rm+sn}=b.$$

However, I'm worried that if $r$ or $s$ happen to be negative then $a^{rm}, a^{sn}$, etc may not make sense, and moreover, I don't see where the fact that I'm working in a domain comes into play. How can this be remedied?

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## marked as duplicate by Arturo Magidin, tomasz, Jennifer Dylan, Alex Becker, Jyrki LahtonenAug 20 '12 at 7:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 4 Answers

That works as long as you pass to the fraction field. But using fractions, the proof is much simpler: excluding the trivial case $\rm\,b=0,\,$ we have $\rm\:(a/b)^m = 1 = (a/b)^n\:$ hence the order of $\rm\,a/b\,$ divides the coprime integers $\rm\,m,n,\,$ thus the order must be $1.\,$ Therefore $\rm\,a/b = 1,\,$ so $\rm\,a = b.\,$

For a proof avoiding fraction fields see this proof that I taught to a student. Conceptually, both proofs exploit the innate structure of an order ideal. Often hidden in many proofs in elementary number theory are various ideal structures, e.g. denominator/conductor ideals in irrationality proofs. Pedagogically, it is essential to bring this structure to the fore.

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Thanks! I'm familiar with fraction fields, so I find this proof quite nice and simple. – hmIII Jun 25 '12 at 20:17

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If $a=0$ or $b=0$, the conclusion follows, so we may assume $a\neq 0$ and $b\neq 0$.

Suppose that $s\lt 0$ (in which case $r\gt 0$). Write $s=-t$ with $t\gt 0$. Then $rm = 1+tn$. So we have $$aa^{tn} = a^{1+tn} = a^{rm} = (a^m)^r = (b^m)^r = b^{rm} = b^{1+tn} = bb^{tn}.$$ Since $a^{tn} = (a^n)^t = (b^n)^t = b^{tn}$, we conclude from $aa^{tn}=bb^{tn}$ that $a=b$.

A symmetric argument holds if $r\lt 0$.

(Basically, we are going to the field of fractions and then clearing denominators "behind the scenes").

Alternatively, say $m = qn+r$, $0\leq r\lt n$. Then $a^ra^{qn} = b^rb^{qn}=b^ra^{qn}$, which yields $a^r=b^r$; so you can replace $m$ with its remainder modulo $n$. Repeating as in the Euclidean Algorithm, we get that if $a^n=b^n$ and $a^m=b^m$, then $a^{\gcd(n,m)} = b^{\gcd(n,m)}$.

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Thanks! One question, is it supposed to be $aa^{tn}=a^{1+tn}+\cdots$  in the first string of equalities? – hmIII Jun 25 '12 at 20:15
+1 I like this way of avoiding the field of fractions. Use cancellation law instead! It's valid in an integral domain! Equivalent, of course, but conceptually a bit simpler IMHO. – Jyrki Lahtonen Jun 25 '12 at 20:29
@hmIII: Yes; I had $s$ and tried to fix it, but I forgot that one. – Arturo Magidin Jun 25 '12 at 20:49
– Gone Jun 25 '12 at 21:07

Your concerns address each other :)

You are worried that $r$ and $s$ may be negative, indicating you wish that inverses for $r$ and $s$ exist so that negative powers for them are defined.

But if you are in a commutative domain, you can work in the field of fractions for the domain, where they are defined!

So, as far as I can see, your logic is completely right, in the field of fractions of the domain.

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Hint: Let $d$ be the least positive integer such that $a^d=b^d$. Show that $d|n$ and $d|m$.

This approach will not require $R$ commutative, or even that $R$ have a multiplicative identity, only that it not have zero divisors.

Specifically, use the division algorithm to show that if $n=dq+r$ with $0\leq r<d$. Then if $r>0$, show $a^r = b^r$, contradicting that $d$ was the least example.

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– Gone Jun 25 '12 at 20:31
@BillDubuque Depends on the level of the reader, I guess.  Sometimes it is nice to emphasize the absolutely elementary nature of a thing, before piling on the abstractions.  For example, this proof works for semi-groups with cancellation, even. – Thomas Andrews Jun 25 '12 at 20:36
– Gone Jun 25 '12 at 20:49
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