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

Let $D$ be an integral domain and $a,~b \in D$. Suppose that $a^n=b^n$ and $a^m=b^m$ for any two some $m,~n$ such that $(m,n)=1$. Prove that $a=b$.

I know that $ab≠0$ since $D$ contains no divisors of zero, but I don’t have an idea as to how to prove this.

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## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 23 at 5:46

As was pointed out by Henning Makholm, the question, as currently stated, has a very short answer (let $m=n=1$). I will assume that the real problem is as follows. Suppose there exist relatively prime positive integers $m$ and $n$ such that $a^m=b^m$ and $a^n=b^n$. Show that $a=b$.

If one of $a$ or $b$ is the zero-element, then the other must be, and therefore $a$ is trivially equal to $b$. So we may assume that neither $a$ nor $b$ is the zero-element. =n=1

Let $d$ be the smallest positive integer such that $a^d=b^d$. We will show that $d=1$, which proves the result.

By dividing $m$ by $d$, we can express $m$ as $qd+r$, where $0\le r<d$. So $a^{qd+r}=b^{qd+r}$, and therefore $$(a^d)^qa^r=(b^d)^q b^r.$$ Since $a^d=b^d$, by cancellation we get $a^r=b^r$. If $r\ne 0$, this contradicts the definition of $d$.

So $r=0$, and therefore $d$ divides $m$. Similarly, $d$ divides $n$. Since $m$ and $n$ are relatively prime, it follows that $d=1$.

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$1$ and $232372$ are relatively prime, so by your assumptions we have $a^1=b^1$ and $a^{232372}=b^{232372}$.

(Did you mean "Suppose that there exist relatively prime $m$ and $n$ such that $a^n=b^n$ and $a^m=b^m$"?)

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You should've used Graham's number! :) – Bruno Joyal Feb 4 '12 at 21:11

Well, you know that $a^{kn}=b^{kn}$ for any positive integer $k$, and $a^{lm}=b^{lm}$ for any positive integer $l$. Multiplying the two equations, $a^{kn+lm}=b^{kn+lm}$.

Since $m$ and $n$ are relatively prime, there exist $k$ and $l$ s.t. $kn+lm\equiv 1$ mod $mn$ (choose $k$ to be the multiplicative inverse of $n$ mod $m$, and similarly for $l$).

Then we can replace $b^{kn+lm}$ with $b*(b^{kn+lm-1})=b*(b^m)^{(kn+lm-1)/m}=b*(a^m)^{(kn+lm-1)/m}$ $=b*a^{kn+lm-1}$.

This gives us $a^{kn+lm}=b*a^{kn+lm-1}$, and canceling the $a$'s gives $a=b$.

Cheers,

Rofler

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This argument does not work as is because one of $k$ or $l$ must be negative. However it can be salvaged by writing $kn = lm+1$ for positive $k$ and $l$. – lhf Feb 4 '12 at 18:56

Clear if $\rm\,a=0\,$ or $\rm\,b=0.\,$ Else the set $\rm\,S\,$ of $\rm\,k \in \mathbb Z\,$ with $\rm a^k = b^k\,$ is closed under subtraction $>0,\,$ i.e.  if $\rm\,j > k\in S\,$ then cancelling $\rm \,0\ne a^k = b^k\,$ from $\rm\,a^j = b^j$ yields $\rm\,a^{j-k} = b^{j-k},\,$ so $\rm\,j\!-\!k\in S.$

Thus by this basic theorem every positive element of $\rm\,S\,$ is a multiple of the least positive $\rm\:d\in S$.

Thus $\rm\:n,m\in S\ \Rightarrow\ d\ |\ n,m\ \Rightarrow\ d\ |\ (n,m) = 1.\:$ Therefore $\rm\: d = 1\in S,\:$ i.e. $\rm\:a^1 = b^1.\ \$ QED

Note $\$ The key idea exploited above is that a set $\rm\,S\,$ of integers closed under $\rm\,subtraction\,$ is also closed under $\rm\,gcd\,$ (hence $\rm\,S\,$ contains $1$ if it contains coprime elements). This is the key idea behind the ancient $\rm\,subtractive\,$ form of the Euclidean algorithm for the $\rm\,GCD.$

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