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Thread: Relitivly prime, unique divisors of divisors

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    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 $\displaystyle d_1d_2=d, d_1|a, d_2|b$
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Chris11 View Post
    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 $\displaystyle 4=2\cdots 2$.
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    @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.
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    Quote Originally Posted by Chris11 View Post
    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 $\displaystyle \gcd(a,d)$.
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