- #1
ihggin
- 14
- 0
Let R be a Principal Ideal Domain. Let a,b \in R, at least one of which is non-zero. Determine precisely all solutions (s,t), where s,t \in R, of the equation sa+tb= \gcd (a,b).
My attempt: since \gcd (a,b) divides both a and b, we can divide both sides by \gcd (a,b) to get the equation s \alpha + t \beta = 1, where \gcd (\alpha , \beta)=1. Just playing around in the integers, if we find a solution x,y, we can generate other solutions by taking x+n \beta, y-n \alpha. However, I'm unsure of a couple of things: 1) how do you show that this gives all solutions (if it does), and 2) how do you find an initial solutions x,y? Of course, any other approaches would be nice to see as well.
Thanks.
My attempt: since \gcd (a,b) divides both a and b, we can divide both sides by \gcd (a,b) to get the equation s \alpha + t \beta = 1, where \gcd (\alpha , \beta)=1. Just playing around in the integers, if we find a solution x,y, we can generate other solutions by taking x+n \beta, y-n \alpha. However, I'm unsure of a couple of things: 1) how do you show that this gives all solutions (if it does), and 2) how do you find an initial solutions x,y? Of course, any other approaches would be nice to see as well.
Thanks.
