Our instructor has given us this problem:
If $ax + by = 8$, what is $\operatorname{gcd}(a, b)$?
I'm confused. Is it not just $8$? Since, say, $\operatorname{gcd}(a, b) = n$, so there must exist some $x, y$ such that $n= ax + by$.
Do I need to prove that?
Assuming that $a,b,x,y\in\Bbb Z$, the only thing that can be guaranteed is that $\gcd(a,b)$ divides $8$.
For example, for $a=b=1$, we have $$4a+4b=8$$ or $$5a+3b=8$$
What Bezout's identity says is:
If $\gcd(a,b)=d$ then there exist two integers $x,y$ such that $ax+by=d$.
and what you have wrongly applied is
If there exist two integers $x,y$ such that $ax+by=d$ then $d=\gcd(a,b)$.
The following statement is true, nevertheless:
If $ax+by=d$ then $\gcd(a,b)$ divides $d$.
