if $d=\gcd(n,a)$ and $n=db,a=dc $ then prove that $\gcd(b,c)=1$

Question

The above snap from Dummit and Foote book pg no. 57 suggests this result. Any explanation of the same will be appreciated.

Answer 1

Apply bezout's formula, $$kn+la=d$$ $$\Longrightarrow kb+lc=1$$

Answer 2

You have $\gcd (n, a) =d$ and we know that by the property of GCD, $d$ must be the largest number that divides both $a$ and $n$.

Consequently, $$\gcd (\frac{n}{d}, \frac{a}{d}) = 1$$

Here, $\frac{n}{d} = b$ and $\frac{a}{d} = c$.

Answer 3

Since we are given $G.C.D(n,a) = d$, We know that $d$ is the highest common divisor to both $a$ and $n$. Now we substitute $n = db$ and $a = dc$ and show that $G.C.D(b,c) = 1$ because if $b$ and $c$ had any common divisor bigger than $1$ , then it should also divide $a$ and $n$ which contradicts the fact that $d$ is the biggest divisor