Is Bezout's lemma enough to confirm the GCD of a number. Is Bezout's lemma enough to confirm the GCD of a number?
So suppose we have $$ax+by=z$$ does this mean $$\gcd(a,b)=z$$
 A: I note that $\mathrm{hcf}$ is usually denoted as $\gcd$ (greatest common divisor). Now from
$$ax+by = z$$
we can only see that $\gcd(a,b) \mid z$ (i.e. $z$ is a multiple of $\gcd(a,b)$) and by Bezout's lemma there exist some $(x,y)$ such that $z=\gcd(a,b)$.
You can't infer anything else about $\gcd(a,b)$ from this, not even if you force $\gcd(x,y)=1$. For example $x=y=1$ gives $z=a+b$. There is not much you could say about the $\gcd$ of some numbers only knowing their sum.
A: Consider the following:
$$ 1\cdot 2 + 1\cdot 3 = 5,$$
and so 
$$\gcd (1,1) = 5.$$
You might try almost any combination of numbers and check for yourself the validity of your statement.
A: A common divisor $\,c\mid a,b\,$ is greatest if it can be written in linear form $\,c = j a + k b,\,$ since then $\,d\mid a,b\,\Rightarrow\, d\mid ja+kb = c,\,$ so $\, d\le c.\,$ Conversely, by Bezout, we know every greatest common divisor can be expressed as such a linear combination.
But only the least positive such linear combination of $\,a,b\,$ equals the gcd. Indeed, such linear combinations are preserved by scaling by arbitrary integers, so any multiple of the gcd is also such a linear combination. Further, one can show that the integers of the form $\ ja + kb \ $ are precisely all integer multiples of the gcd, i.e. $\ a \Bbb Z + b\Bbb Z\, =\, \gcd(a,b)\Bbb Z.\,$ Said equivalently, there exist integers $\,x,y\,$ such that $\ a x + by = c \iff\gcd(a,b)\mid c,\,$ see here.
