- #1
ihggin
- 14
- 0
Let [tex]R[/tex] be a Principal Ideal Domain. Let [tex]a,b \in R[/tex], at least one of which is non-zero. Determine precisely all solutions [tex](s,t)[/tex], where [tex]s,t \in R[/tex], of the equation [tex]sa+tb= \gcd (a,b)[/tex].
My attempt: since [tex]\gcd (a,b)[/tex] divides both [tex]a[/tex] and [tex]b[/tex], we can divide both sides by [tex]\gcd (a,b)[/tex] to get the equation [tex]s \alpha + t \beta = 1[/tex], where [tex]\gcd (\alpha , \beta)=1[/tex]. Just playing around in the integers, if we find a solution [tex]x,y[/tex], we can generate other solutions by taking [tex]x+n \beta, y-n \alpha[/tex]. 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 [tex]x,y[/tex]? Of course, any other approaches would be nice to see as well.
Thanks.
My attempt: since [tex]\gcd (a,b)[/tex] divides both [tex]a[/tex] and [tex]b[/tex], we can divide both sides by [tex]\gcd (a,b)[/tex] to get the equation [tex]s \alpha + t \beta = 1[/tex], where [tex]\gcd (\alpha , \beta)=1[/tex]. Just playing around in the integers, if we find a solution [tex]x,y[/tex], we can generate other solutions by taking [tex]x+n \beta, y-n \alpha[/tex]. 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 [tex]x,y[/tex]? Of course, any other approaches would be nice to see as well.
Thanks.