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Math Help - if ra\equiv rb (mod m), then a\equiv b (mod \frac{m}{gcd(r,m)})

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    if ra\equiv rb (mod m), then a\equiv b (mod \frac{m}{gcd(r,m)})

    if ra\equiv rb \ (\text{mod} \ m), then a\equiv b \ \left(\text{mod} \  \frac{m}{gcd(r,m)}\right)

    Let d=gcd(r,m)
    Then d|r \ \text{and} \ d|m

    \Rightarrow ds=r \ \text{and} \ dt=m, \ \ st,\in\mathbb{Z}

    t=\frac{m}{gcd(r,m)}

    m|[r(a-b)]\Rightarrow t|[r(a-b)]

    How can show t and r are coprime so t|(a-b)?
    Last edited by dwsmith; June 15th 2011 at 08:52 AM.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: if ra\equiv rb (mod m), then a\equiv b (mod \frac{m}{gcd(r,m)})

    Quote Originally Posted by dwsmith View Post
    if ra\equiv rb \ (\text{mod} \ m), then a\equiv b \ \left(\text{mod} \  \frac{m}{gcd(r,m)}\right)

    Let d=gcd(r,m)
    Then d|r \ \text{and} \ d|m

    \Rightarrow ds=r \ \text{and} \ dt=m, \ \ st,\in\mathbb{Z}

    t=\frac{m}{gcd(r,m)}

    m|[r(a-b)]\Rightarrow t|[r(a-b)]

    How can show t and r are coprime so t|(a-b)?



    d=gcs(r,m)
    ra\equiv rb(\mod m)
    -----------------
    d=gcd(r,m)

    From the given we can write:

    (1)  r(a-b)=ra-rb=lm, for k\in\mathbb{Z}.


    d=gcd(r,m), therefor there are exist p,q co-prime. so that: m=dp, r=dq, we put these in (1), and we will get:

    q(a-b)=kp

    Hence, p|q(a-b) and gcd (p,q)=1.

    Recalling the Euclid's lemma we will get: p|(a-b) or a\equiv b(\mod p) or:


    a\equiv b(\mod \frac{m}{d})
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