1. ## GCD Question

The problem:
Let a,b,n be integers. n >0, such that n | (a-b) [n divides (aib)].
Show that (n,a) = (n,b) [(x,y) <==> the GCD of x and y].

Definitions
a | b means that b = ak + r for some integer r,k k=0;

Lemmas
Let a,b be integers. If a = bq+r, then (a,b) = (b,r).

Proof:
Since n | (a-b)
a-b = nk for some integer k.
a = nk + b
Using Lemma, (a,n) = (n,b).

I'm just wondering if there are any flaws in this proof, given I can use the lemma? I just want to make sure, seems a bit easy but my teacher is very strict and was just wondering if any of you could spot some possible criticisms.

Thanks

2. Originally Posted by Th3sandm4n
The problem:
Let a,b,n be integers. n >0, such that n | (a-b) [n divides (aib)].
Show that (n,a) = (n,b) [(x,y) <==> the GCD of x and y].

Definitions
a | b means that b = ak + r for some integer r,k k=0;

Lemmas
Let a,b be integers. If a = bq+r, then (a,b) = (b,r).

Proof:
Since n | (a-b)
a-b = nk for some integer k.
a = nk + b
Using Lemma, (a,n) = (n,b).

I'm just wondering if there are any flaws in this proof, given I can use the lemma? I just want to make sure, seems a bit easy but my teacher is very strict and was just wondering if any of you could spot some possible criticisms.

Thanks
Look at the first problem here.

3. Yeah that is a more rock solid proof.

Just out of curiosity though, was there anything wrong in the way I used the Lemma or my manipulation? Did it prove it?
I always get really excited when I seem to figure out a good way to use things, and was just wondering if I was correct in thinking that?

4. Originally Posted by Th3sandm4n
Yeah that is a more rock solid proof.

Just out of curiosity though, was there anything wrong in the way I used the Lemma or my manipulation? Did it prove it?
I always get really excited when I seem to figure out a good way to use things, and was just wondering if I was correct in thinking that?