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I did:

$$ca = cb * k + r \Leftrightarrow \\ ca - cb*k = r \\ c(a-bk)=r \\ a-bk = r/c \\ a = bk +r/c$$

So, $gcd(a,b) = r/c$

What do I do next?

Ethan
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SilenceOnTheWire
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    Also: https://math.stackexchange.com/q/1437530/42969, https://math.stackexchange.com/q/3430009/42969, https://math.stackexchange.com/q/982040/42969, https://math.stackexchange.com/q/705862/42969 – Martin R Nov 11 '19 at 14:37
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    You asked the same question before: https://math.stackexchange.com/q/3430009/42969 – Martin R Nov 11 '19 at 15:42

1 Answers1

-1

Hint:

It suffices to show that $|c|\gcd(a,b)$ divides $ca$ and $cb$

and any number that divides $ca$ and $cb$ divides $|c|\gcd(a,b)$.

J. W. Tanner
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  • And how do you do that? – SilenceOnTheWire Nov 11 '19 at 15:20
  • Well, $\gcd(a,b)$ divides $a$ and $b$, and $|c|$ divides $c$, so $|c|\gcd(a,b)$ divides $ca$ and $cb$; on the other hand, if $d=k|c|\gcd(a,b)$ divides $ca$ and $cb$, then $k\gcd(a,b)$ divides $a$ and $b$ and hence $gcd(a,b)$ so $k$ divides $1$ – J. W. Tanner Nov 11 '19 at 16:35