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Heres what I was thinking ...

and its a = b (mod c) .. typoed it

If a = b mod c then b = a + ck
then gcd(b,c) = gcd(a+ck,c) = gcd(a,c)
(I think)

Oh, those arrows go both ways. It's an equality. Woops.

Oh, I had it backwards. It works the same way, though.

I was relying on a theorem (or an extension of one, at least I think I was).

In fact this is an extension of gcd(m,n) = gcd(n,m%n)

id have to see a more detailed version of this: gcd(b,c) = gcd(a+ck,c)

k, a-b = ck like u have...

your just subing in b with a+ck ....

takes me a few times to read it right :)

Yes, relying on the theorem above....(so I guess what you want is a proof of that, in effect...

a-b = ck like u have
Let gcd(a,c) = n -> n divdes a and c so a = nq and c = nr (some q and r)

Hope that's right (lots of letters to keep track of).

Found a prettier version of the whole thing.....