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Modular Arithmetic and Diophantine Equations

  1. Oct 25, 2012 #1
    If one is solving a modular equation:

    [itex]4k \equiv 1 \: (\text{mod } n)[/itex]

    with n even, known, for k, then one needs to find the inverse of 4 modulo n:

    [itex] 4x - 1 = nc [/itex]
    [itex] 4x - nc = 1 [/itex]

    But this only has solutions iif (4,n) = 2 (n is even, but not a multiple of 4), which doesn't divide 1, so there is no inverse of 4 modulo n. Does this mean that there isn't a k that satisfies the original equation?

  2. jcsd
  3. Oct 25, 2012 #2


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    hi drewfstr314! :smile:

    that's right :smile:

    4k is even, and 1 (mod n) is odd (since n is even), so there's no solution …

    what is worrying you about that? :confused:
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