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Math Help - Indices

  1. #1
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    Indices

    1.) Find all the solutions of 4^(x) = 13 (mod 17) using indices.

    Then, let m be a pos. integer w/ a primitive root r. Also, let a be a pos. integer w/ gcd(a,m) = 1. Through 2.) and 3.), you'll investigate the relationship between ind_(r)a and ind_(r)a^(-1) where a^(-1) is the inverse (multiplicative) of 'a' (mod m)

    2.) For each of 3 examples, specify m, r, a. Then find a^(-1), ind_r(a) and ind_(r)a^(-1).

    3.) From #2, find a formula relating ind_(r)a and ind_(r)a^(-1). You don't have to prove this.
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  2. #2
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    Hello, flash101!

    I'm not familar with "using indices", but here's my version of #1 . . .


    1.) Find all the solutions of 4^x .= .13 .(mod 17)

    Examine consecutive powers of 4:

    . . 4^
    0 .= .1 .(mod 17)

    . . 4^
    1 .= .4

    . . 4^
    2 .= -1

    . . 4^
    3 .= 13

    . . 4^
    4 .= .1

    . . 4^
    5 .= .4

    . . 4^
    6 .= -1

    . . 4^
    7 .= 13


    We see that the mod-17 values are cyclic: 1, 4, -1, 13, . . .

    . . and the solutions are: .x = 3, 7, 11, 15, . . .


    That is: .x .= .4k - 1, for any integer k.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, flash101!

    I'm not familar with "using indices", but here's my version of #1 . . .


    Examine consecutive powers of 4:

    . . 4^0 .= .1 .(mod 17)

    . . 4^1 .= .4

    . . 4^2 .= -1

    . . 4^3 .= 13

    . . 4^4 .= .1

    . . 4^5 .= .4

    . . 4^6 .= -1

    . . 4^7 .= 13


    We see that the mod-17 values are cyclic: 1, 4, -1, 13, . . .

    . . and the solutions are: .x = 3, 7, 11, 15, . . .


    That is: .x .= .4k - 1, for any integer k.
    Examples using "indices" include:

    11^(x) = 5 (mod 13)

    r = 2 (primitive root)

    ind_(2)(11^(3x)) = ind_(2)5(mod phi(13))
    3xind_(2)11 = ind_(2)5 (mod 12)
    3x*7 = 9 (mod 12)
    7x = 3 (mod 4)
    x = 1 (mod 4)
    x = 1, 5, 9 (mod 12)


    Another example:

    3^(x) = 2 (mod 23)

    r = 5

    ind_(5)3^(x) = ind_(5)2 = 2 (mod 22)

    ind_(5)3^(x) = x ind_(5)3 = 16x (mod 22)
    16x = 2 (mod 22) Divide by 2:
    8x = 1 (mod 11)
    x = 7 (mod 11)
    x = 7 or 18 (mod 22)
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  4. #4
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    I have a question about the gcd(a,m) in this type problem.
    for example....

    7x^20 = = 34(mod 37) [using indices]
    I (7x^20) = I(34) (mod 36)
    ...
    ...
    ..
    .
    20x = 27(mod 36) here I would take the gcd (20,36) = 4 divided by 27/\ therefore there are no solutions???

    Is this accurate?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by duggaboy View Post
    I have a question about the gcd(a,m) in this type problem.
    for example....

    7x^20 = = 34(mod 37) [using indices]
    I (7x^20) = I(34) (mod 36)
    ...
    ...
    ..
    .
    20x = 27(mod 36) here I would take the gcd (20,36) = 4 divided by 27/\ therefore there are no solutions???

    Is this accurate?
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