TITLE: Converse of Bézout's identity
QUESTION [6 upvotes]: Given that  we know gcd$(a,b)=sa+tb$ where $s,t \in \mathbb Z$, I wonder  is the converse true?
I.e., can we say that the value of $ax+by$ is the gcd of any $2$ of $a,x,b,y$?

REPLY [8 votes]: Yes and no. 
If an integer of the form $as + bt$ is a divisor of $a$ and $b$, then it is a greatest common divisor.
To see this note that a divisor common to $a$ and $b$ will divide $as$ and $bt$ and hence also $as + bt$ for every choice of $s$ and $t$. Thus a GCD will divide every number of the form $as + bt$. Thus, if you can express some common divisor in this form it will be a multiple of a GCD and hence a GCD.
Yet, of course not every integer of this form is a divisor of $a$ and $b$. 
To see this just set $s= 17$ and $t=0$. The $as + bt=17a$ which is clearly not a GCD of $a,b$. (Well, except when both happen to be $0$ but that's a detail.)