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Math Help - Modular Arithmetic- Quick question

  1. #1
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    Modular Arithmetic- Quick question

    Hello

    I've been having difficulty with understanding a concept within modular arithmetic

    The question I have been stuck on for a while now is:

    Given 5x = -2(mod 8)

    find x^2(mod 8)

    My attempt at a solution:

    5x = -2(mod 8)'
    25x^2 = 4 (mod 8)
    x^2 = 4 (mod 8), as 25 = 4 (mod 8)

    The answer at the back says 1, and I'm not sure where I went wrong

    Thanks in advance
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Wandering View Post
    Hello

    I've been having difficulty with understanding a concept within modular arithmetic

    The question I have been stuck on for a while now is:
    i can't find a mistake. for example x=6 satisfies 5x=-2(mod 8) and 6^2=4(mod 8)
    Given 5x = -2(mod 8)

    find x^2(mod 8)

    My attempt at a solution:

    5x = -2(mod 8)'
    25x^2 = 4 (mod 8)
    x^2 = 4 (mod 8), as 25 = 4 (mod 8)

    The answer at the back says 1, and I'm not sure where I went wrong

    Thanks in advance

    i can't find a mistake. for example x=6 satisfies 5x=-2(mod 8) and 6^2=4(mod 8)
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  3. #3
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    So the answer at the back must be wrong?

    Thanks a lot
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Wandering View Post
    Hello

    I've been having difficulty with understanding a concept within modular arithmetic

    The question I have been stuck on for a while now is:

    Given 5x = -2(mod 8)

    find x^2(mod 8)

    My attempt at a solution:

    5x = -2(mod 8)'
    25x^2 = 4 (mod 8)
    x^2 = 4 (mod 8), as 25 = 4 (mod 8)

    The answer at the back says 1, and I'm not sure where I went wrong

    Thanks in advance
    I too agree with your answer, nice going!

    One little thing which I'm presuming is a typo since you got the correct answer. You wrote 25 = 4 (mod 8). In actuality 25 = 1 (mod 8).

    -Dan
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  5. #5
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    Hello, Wandering!

    Sometimes early squaring can introduce extraneous roots.
    So I solved it head-on.




    . .








    . .





    You are correct . . . Good work!

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