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Math Help - if (a,m)=1, then ax\equiv 1 (mod m) has a unique sol.

  1. #1
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    if (a,m)=1, then ax\equiv 1 (mod m) has a unique sol.

    if (a,m)=1, then ax\equiv 1 \ (\text{mod} \ m) has a unique sol.

    By Bezout, there exist x,y\in\mathbb{Z} such that 1=ax+my

    Now, we can write that as:

    my=1-ax\Rightarrow m(-y)=ax-1\Rightarrow ax\equiv 1 \ (\text{mod} \ m)

    Correct?
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  2. #2
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    Re: if (a,m)=1, then ax\equiv 1 (mod m) has a unique sol.

    Hmm...

    I don't know about that. What we are trying to prove is that there is a unique value for x (up to congruence modulo m) satisfying ax\equiv 1\,(\mathrm{mod}\,m) (where a,m are fixed). I don't think that what you have done gives us the uniqueness.

    Try this: suppose there exist x,y such that ax\equiv ay\equiv 1\,(\mathrm{mod}\,m). We try to prove that x\equiv y\,(\mathrm{mod}\,m).

    It's fairly simple: what happens if you multiply the congruence ax\equiv 1\,(\mathrm{mod}\,m) by y?
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