## Solve linear Diophantine equations in two variables in a Euclidean domain

Let be a Euclidean Domain.

- Prove that if and divides , then divides . More generally, show that if divides with , then divides .
- Consider the Diophantine Equation , where , , and are integers and . Suppose is a solution- that is, . Prove that the full set of solutions to this equation is given by and as ranges over the integers.

- Suppose first that and that . By Theorem 4, there exist such that . Then . Since , we have for some . So , and thus . So .
More generally, say , , and . Again by Theorem 4, we have such that . Since , we have for some . Now , so that , and thus Since is an integral domain, , and thus divides .

- Write , and suppose is a solution to . Then , so that . Since divides , we have that divides . Say ; then . Substituting into , we see that , so that . Since divides and is a domain, we have .
Moreover, it is straightforward to show (as we did in this previous exercise) that every pair of this form is in fact a solution. Thus we have completely characterized the solutions of .

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## Comments

In part 2, isn’t it that a(x – x_0) = b(y_0 – y), rather than a(x – x_0) = b(y – y_0)?

Thanks!

please send for me revision questios..

send for me solved examples

It’s not too hard to make up examples- just pick any numbers for , , , and .

For instance, with and , the equation has as a solution. Every other solution has the form where is an arbitrary integer.