# 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 Lahtonen♦Aug 20 '12 at 7:24

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

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
@Jyrki It's really nothing but the (subtractive form of the) Euclidean algorithm for the GCD - see the explanation here. It's quite trivial when viewed this way. –  Bill Dubuque Jun 25 '12 at 21:07

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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That's essentially the same as the proof that I linked to in my answer. However, I think it is important to emphasize the innate ideal structure. –  Bill Dubuque 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
The linked proof also works there. Did you read it? If you do you'll find that there is no "piling on of abstraction". I've had success teaching such proofs to bright high-school students. –  Bill Dubuque Jun 25 '12 at 20:49