Converse of Bézout's identity

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 $a,x,b,y$?

Solution 1:

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.)

Solution 2:

Clearly not, since doubling $x$ and $y$ doubles $ax+by$. Bézout’s result says that $\gcd(a,b)$ is the smallest positive integer of the form $ax+by$ for $x,y\in\Bbb Z$, and that all other numbers of this form are integer multiples of $\gcd(a,b)$.