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

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- Nov 23rd 2010, 06:58 PMChris11Relitivly 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 $\displaystyle d_1d_2=d, d_1|a, d_2|b$ - Nov 23rd 2010, 08:07 PMDrexel28
- Nov 24th 2010, 07:24 AMPetek
@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. - Nov 24th 2010, 08:40 AMOpalg
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 $\displaystyle \gcd(a,d)$.