Relitivly prime, unique divisors of divisors

**Chris11** · November 23rd 2010, 06:58 PM

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

**Drexel28** · November 23rd 2010, 08:07 PM

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 $4=2\cdots 2$ .

**Petek** · November 24th 2010, 07:24 AM

@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.

**Opalg** · November 24th 2010, 08:40 AM

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 $\gcd(a,d)$ .

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