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

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.