# Relitivly prime, unique divisors of divisors

• November 23rd 2010, 06:58 PM
Chris11
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 $d_1, d_2$ such that $d_1d_2=d, d_1|a, d_2|b$
• November 23rd 2010, 08:07 PM
Drexel28
Quote:

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 $d_1, d_2$ such that $d_1d_2=d, d_1|a, d_2|b$

This makes no sense. $(4,17)=1$ and surely $4\mid 4\cdot 17$ but $d=4\cdot 1$ and $4=2\cdots 2$.
• November 24th 2010, 07:24 AM
Petek
@Drexel28 - In your example, we have a = 4, b = 17 and d = 4. If we let $d_1 = 4$ and $d_2 = 1$, 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.
• November 24th 2010, 08:40 AM
Opalg
Quote:

Originally Posted by Chris11
Let (a, b)=1. let d|ab. show that there exist unique $d_1, d_2$ such that $d_1d_2=d, d_1|a, d_2|b$

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

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