anonymous
  • anonymous
let \(a,b,c\in\mathbb{Z}\), \(\gcd(a,b)=1\), and \(c|(a+b)\). how can i show that \(\gcd(c,a)=\gcd(c,b)=1\)?\[\]i need some pointers.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Its very easy..... You need to show that : \[ (x_{0},y_{0}) \in \mathbb{Z} c x_{1} + a y_{1} = c x_{2} + b y_{2} = 1\]
anonymous
  • anonymous
We will proceed as follows : Given: c divides (a + b) therefore for some integers y1 and y2 c divides ay1 + by2 .........(1) i.e. ay1 + by2 = ck for some integer k. Now k being an integer we can write it as a difference of two more integers, say k = x2 - x1 . Thus we now have : ay1 + by2 = c(x2 - x1) But (1) is nothing but 1 as (a,b) = 1. Therefore, ay1 + by2 = c (x2 - x1 ) = 1 or, ay1 + cx1 = cx2 + by2 = 1. QED
anonymous
  • anonymous
that makes perfect sense. thank you very much

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anonymous
  • anonymous
Welcome.

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