Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $D$ be an integral domain and let $a^m=b^m$ and $a^n=b^n$ where $m$ and $n$ are relatively prime integers, $a,b \in D$.

How do I show $a=b$?

share|improve this question
Hint: If $m$ and $n$ are relatively prime, then there exist integers $x$ and $y$ such that $mx + ny = 1$. –  ShreevatsaR Nov 16 '11 at 13:31
Can you show this in the special case where $m = n+1$? –  Hans Engler Nov 16 '11 at 13:31
Subhint: let $D$ the set of $k$ such that $a^k=b^k$. You know that $n$ and $m$ are in $D$ and you want to show that $1$ is in $D$. What other integers do you know as being in $D$? –  Did Nov 16 '11 at 13:44
Sorry: not your $D$... –  Did Nov 16 '11 at 14:49
maybe you could use this identity : (math.stackexchange.com/q/7473/15660) –  pedja Nov 16 '11 at 14:49

4 Answers 4

up vote 1 down vote accepted

No generality is lost by supposing $m < n$. So $a^n=b^n$ implies $a^{m+(n-m)}=b^{m+(n-m)}$, or $a^ma^{n-m}=b^mb^{n-m}$. In integral domains, there's a cancellation property, so $a^{n-m}=b^{n-m}$.

The pair $(m,n)$ has now been replaced by the pair $(m,n-m)$. If you iterate that process, replacing the pair you've got with the pair consisting of the larger of the two and the difference---the larger minus the smaller, that's Euclid's algorithm. It ends when you reach the gcd.

share|improve this answer
Here the integral domain doesn't have a unit. Will the cancellation property still hold? –  Mohan Nov 16 '11 at 14:25
@user Suppose $xy = xz$ with $x \ne 0$. Then $x(y-z)=0$; by the definition of an integral domain, since $x \ne 0$, it must be the case that $y - z = 0$. –  Srivatsan Nov 16 '11 at 14:34
@user774025 It's been a while since I've thought these issues through, but I've just look in Herstein's undergraduate text, Topics in Algebra, and he gives a definition of integral domain that doesn't assume that those have a unit, and in that context proves that every integral domain is embedded in a field of quotients. If it's embedded in a field of quotients, then it must have the cancellation property. –  Michael Hardy Nov 17 '11 at 16:04

Since $m$ and $n$ are coprime then $x m - y n=1$ for some $x,y \in \mathbb Z$. The equality $a^m=b^m$ implies that $a^{xm}=b^{xm}$ so $a^{1+yn}=b^{1+yn}$ which implies that : $$(*)\;\;\; \;a a^{yn}=b b^{yn}$$ Since $a^{n}= b^{n}$ then $a^{yn}=b^{yn}\not = 0$ so we can cancel $a^{yn}$ and $b^{yn}$ from both sides of $(*)$ since we are working in an integral domain, and then we get $a=b$.

share|improve this answer
What if $y \lt 0$? then the expression $ a^{yn}$ will be meaningless? –  Mohan Nov 16 '11 at 15:48
It has meaning in the quotient field of the domain $D$. –  Andrea Nov 19 '11 at 13:29

Hint $\ $ The set $\,S\,$ of naturals $\,k>0\,$ such that $\,a^k = b^k\,$ is nonempty and closed under positive subtraction, i.e. $\, j>k \in S\,\Rightarrow\,j-k\in S\,$ (by cancelling $\,a^k = b^k\,$ from $a^j = b^j).$ Therefore, by a fundamental lemma, the least element $\,\ell\in S\,$ divides every element of $\,S.\,$ In particular, $\,\ell\,$ is a common divisor of the coprimes $\,m,n\in S,\,$ thus $\,\ell = 1.\,$ Hence $\,1\in S\,\Rightarrow\, a^1 = b^1.\ \ $ QED

share|improve this answer

What's the necessity of $D$ to be a domain? As $(m,n) = 1$ we have two integers $x$ and $y$ such that $mx+ny = 1$

so, $a^{mx} = b^{mx}$ and $a^{ny} = b^{ny}$ which implies, $a^{mx+ny} = b^{mx+ny} = a = b$

share|improve this answer
Your proof only works if $x,y$ are both positive. –  Martin Brandenburg Mar 2 at 15:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.