Bezout's theorem extension (regarding uniqueness of x,y and converse).

Question

In my book Bezout's theorem is given as:

If a,b are integers, not both zero, then GCD(a,b) exists and there exist integers x and y such that (a,b)=ax +by.

I tried to prove it in 4 steps:

1.Prove the existence of GCD.

2.Prove the uniqueness of GCD (There is only one GCD).

3.Prove that (a,b)=ax +by.

4.Prove that x and y are not unique (There can be many x and y which satisfy (a,b)=ax +by).

5.Prove the converse (As a question in my book was solved using the converse of Bezout's theorem).

I proved the first three parts (though i had to take a hint from the book), but i am stuck at last two and I have no idea how to approach them (4th and 5th).

Any suggestion is heartily welcome.

So for 4 you mean to say that x and y are not unique, right? — Bram28, Nov 08 '16 at 11:35
More succinctly Bezout's gcd equality may be expressed as $\ a\Bbb Z + b\Bbb Z = (a,b)\Bbb Z,,$ e.g. see here.. This structural forumulation will become clearer when you learn about groups and ideals in abstract algebra. — Bill Dubuque, Nov 08 '16 at 18:35

score 2 · Accepted Answer · answered Nov 08 '16 at 12:10

For 4: if $(x,y)$ is a solution to $gcd(a,b) = ax + by$, then $(x+nb,y-na)$ is a solution as well for any whole number $n$.

For 5: (reverse Bezout) Note that it is not true that if $z = ax + by$, then $z = gcd(a,b)$. It is true, however, that if $z = ax + by$, then $z = n*gcd(a,b)$ for some whole number $n$, i.e. $z$ will be a multiple of $gcd(a,b)$. And that is straightforward, since both $a$ and $b$ are multiples of $gcd(a,b)$, so $ax$ and $by$ are as well, and hence $ax + by$ is.

Bezout's theorem extension (regarding uniqueness of x,y and converse).

1 Answers1