What is variable c in Bezout's Identity

Question

I'm trying to learn Carmichael's function from this answer.

I analyzed this rule:

If $\gcd(a,b) = 1$ and $a|bc$, then $a|c$.

If greatest common divisor of numbers a and b is 1 (if they are coprime) and a is divisible by b multiplied by c, then a is divisible by c.

But what is this c?

I thought it was the output of the gcd function. Thus, I substituted variables with prime numbers (where a=5, b=3, c=gcd(a,b)) and got this:

If gcd(5, 3) = 1 and 5|3(1), then 5|1.

Now obviously, the statement above is incorrect since 5 is not evenly divisible by 3 and yet, 5 is divisible by 1.

Then, the author of the answer mentioned Bezout's Identity -

$a$ and $b$ are coprime if and only if there exist $x$ and $y$ such that $ax+by=1$. Multiplying through by $c$ we get $c = acx + bcy$. Since $a$ divides both $acx$ and $bcy$ (the latter by virtue of dividing $bc$), then $a$ divides $c$.

But, yet, how is variable c obtained?

Thank you!

Answer 1

If you are intersted to see how $c$ is linked to Bezout's theorem, consider that since: $$a|bc\implies bc=ak $$

Since also: $$\gcd(a,b)=1$$

for Bezout's theorem $\exists x,y \in \mathbb{Z}$ such that:

$$ax+by=1 \implies acx+bcy=c \implies acx+aky=c \implies a(cx+ky)=c$$

Thus $$c|a \quad \square$$

Answer 2

What the statement means is this: if $a$, $b$, and $c$ are integers, if $a$ and $b$ are co-prime, and if $a$ divides $bc$, then $a$ divides $c$.