# Thread: Relitivly prime, unique divisors of divisors

1. ## Relitivly prime, unique divisors of divisors

I was wondering if someone could help me prove the following.

Let (a, b)=1. let d|ab. show that their exists unique $\displaystyle d_1, d_2$ such that $\displaystyle d_1d_2=d, d_1|a, d_2|b$

2. Originally Posted by Chris11
I was wondering if someone could help me prove the following.

Let (a, b)=1. let d|ab. show that their exists unique $\displaystyle d_1, d_2$ such that $\displaystyle d_1d_2=d, d_1|a, d_2|b$
This makes no sense. $\displaystyle (4,17)=1$ and surely $\displaystyle 4\mid 4\cdot 17$ but $\displaystyle d=4\cdot 1$ and $\displaystyle 4=2\cdots 2$.

3. @Drexel28 - In your example, we have a = 4, b = 17 and d = 4. If we let $\displaystyle d_1 = 4$ and $\displaystyle d_2 = 1$, then it seems to me that we've met the requirements of the problem.

@Chris11 - Are you allowed to use the Fundamental Theorem of Arithmetic (i.e., unique factorization) in your solution? If so, that's a hint.

4. Originally Posted by Chris11
Let (a, b)=1. let d|ab. show that there exist unique $\displaystyle d_1, d_2$ such that $\displaystyle d_1d_2=d, d_1|a, d_2|b$
You could start by letting $\displaystyle d_1 = \gcd(a,d)$. Then $\displaystyle d_1$ divides $\displaystyle d$, so let $\displaystyle d_2 = d/d_1$. Also, there exist integers $\displaystyle x$, $\displaystyle y$ such that $\displaystyle ax+dy = d_1$. Multiply that equation by $\displaystyle b$, and use the fact that $\displaystyle ab$ is a multiple of $\displaystyle d$ to conclude that $\displaystyle d_2$ divides $\displaystyle b$.

For the uniqueness, you could try to show that $\displaystyle d_1$ must necessarily be equal to $\displaystyle \gcd(a,d)$.