Bézout identity and the structure of solutions

[safwane]

Bézout's identity — Let a and b be integers with greatest common divisor d. Then there exist integers x and y such that ax + by = d. More generally, the integers of the form ax + by are exactly the multiples of d.

In this page:

Bézout's identity - Wikipedia

en.wikipedia.org

The author says the following:

If a and b are not both zero and one pair of Bézout coefficients (x, y) has been computed (e.g., using extended Euclidean algorithm), all pairs can be represented in the form:

where k is an arbitrary integer, d is the greatest common divisor of a and b, and the fractions simplify to integers.

I try to prove that all pairs can be represented in the above form and no pair exist outside this form. But I am not able to find the right idea.

 

[blamocur]

A couple of recommendations:

• Replace b/d and a/d by, say, A and B so you don't have to deal with fractions. A and B are coprime, and [imath]Ax + By = 1[/imath].
• For any other pair [imath]x_2, y_2[/imath] such that [imath]ax_2+by_2=1[/imath] subtract one equation from another.

 

[safwane]

A couple of recommendations:

• Replace b/d and a/d by, say, A and B so you don't have to deal with fractions. A and B are coprime, and [imath]Ax + By = 1[/imath].
• For any other pair [imath]x_2, y_2[/imath] such that [imath]ax_2+by_2=1[/imath] subtract one equation from another.

Thank you very much for this clarification for proof

 

[Steven G]

Can we see your proof??