Exploring the Consequences of c^a=d^a & c^b=d^b in a Domain

[Metric_Space]

Homework Statement

Let c and d be two non-zero elements of a domain D. If a and b are integers s.t gcd(a,b)=1, a>0, b > 0.

If we know c^a=d^a and c^b=d^b,

does it follow that c=d?

Homework Equations

The Attempt at a Solution

I'm thinking divinding the two might be pointless ..but subtracting might be a good idea.
c^a-c^b=d^a-d^b.

Somehow have to use the fact that we are working in a domain? The hint is to consider the Quot(D) and work in it.

 

[Metric_Space]

actually...maybe I'm wrong, now I'm thinking to consider

(c^a)/(d^a)=(c^b)/(d^b) (given since non-zero) ...

I'm thinking if one messes around with this ...I'll get the answer. Does this seem on the right track?

 

[Metric_Space]

I'm getting then

(c^a)*(d^b)-(c^b)*(d^a)=0

I sbustituted in c^b=d^b to get

(c^a)*(d^b)-(d^b)*(d^a)=0

but that gets me back to where I started...

 

[micromass]

Hi Metric Space!

Do you know Bezout's theorem? The one that says that if gcd(a,b)=1, then there exist integers k and l such that ka+lb=1.

I suggest you use this.

 

[Metric_Space]

interesting...I'll give it a shot -- I know the result but not the name...until now

 

[micromass]

Well, actually, Bezout's theorem says that for each a and b, there exists k and l such that

$$ka+lb=gcd(a,b)$$

I just applied it here with gcd(a,b)=1.

It's one of the more important theorems in abstract algebra and number theory. It's very useful!

 

[Metric_Space]

care to give another hint? I plugged that into one of my equations and just got a big mess

 

[micromass]

Complete this:

$$c=c^{ak+lb}=...$$

 

[Metric_Space]

ah...very useful hint

I solved for a before a in ak+lb=1...and that's what made a mess ...this hint 'solved' it - thanks!