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amistre64
Group Title
Show that if a,b,c are integers; c>0; and
a [congruent] b(mod m), then gcd(a,c) = gcd(b,c)
 2 years ago
 2 years ago
amistre64 Group Title
Show that if a,b,c are integers; c>0; and a [congruent] b(mod m), then gcd(a,c) = gcd(b,c)
 2 years ago
 2 years ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Heres what I was thinking ...
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
and its a = b (mod c) .. typoed it
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
since a = b (c); then a = b + ck, for some integer k a  b = ck ax1 + bx2 = ck ax1 + bx2 = c(y1+y2) ax1 + bx2 = cy1 + cy2 ax1 + cy1 = bx2 + cy2 ax1 + cy1 = g1, for some integer g g1 = bx2 + cy2 gcd(a,c) = gcd(b,c)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
If a = b mod c then b = a + ck then gcd(b,c) = gcd(a+ck,c) = gcd(a,c) (I think)
 2 years ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
That's true. You might want to give more reasoning though: Let d = gcd(b,c). Then, gcd(b,c) => bd and cd. => a+ckd and cd. Since cd, ckd. And, since ckd, a+ckd => ad. And, then you're done. Easier to see this way, I think.
 2 years ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
Oh, those arrows go both ways. It's an equality. Woops.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
(b,c) = d therefore db and dc there is some thrm that states that the gcd of 2 numbers can be written as a linear combination of those numbers d = bx + cy
 2 years ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
Oh, I had it backwards. It works the same way, though.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
I was relying on a theorem (or an extension of one, at least I think I was).
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
In fact this is an extension of gcd(m,n) = gcd(n,m%n)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
id have to see a more detailed version of this: gcd(b,c) = gcd(a+ck,c)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
k, ab = ck like u have...
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
your just subing in b with a+ck ....
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
takes me a few times to read it right :)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Yes, relying on the theorem above....(so I guess what you want is a proof of that, in effect...
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
ab = ck like u have Let gcd(a,c) = n > n divdes a and c so a = nq and c = nr (some q and r)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Hope that's right (lots of letters to keep track of).
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
So ab = ck > nq  b = nrk > b = n(qrk) > n divides b Do same for other side and it should drop out (I hope)
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Found a prettier version of the whole thing.....
 2 years ago
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