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Math Help - Prove K^2 mod 4 = 1 ?

  1. #1
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    Prove K^2 mod 4 = 1 ?

    This is my first post

    I have been working on this for far too long

    Can someone let me know if this is correct?

    "k = 2n + 1. Then k^2 (is congruent to) 4n^2 + 4n + 1 (is congruent to) 1 mod 4"

    Should I add more to it? Am I at least on the right track?
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  2. #2
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    Re: Prove K^2 mod 4 = 1 ?

    Hey brianjasperman.

    What you did is correct but if your teacher is picky you should probably write 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1 just to add that extra clarity to the proof.
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  3. #3
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    Re: Prove K^2 mod 4 = 1 ?

    So I would prove the equation by saying:
    "k = 2n + 1. Then 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1"

    OR

    "k = 2n + 1. Then k^2 (is congruent to) 4n^2 + 4n + 1 (is congruent to) 1 mod 4, and 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1"

    Thanks for the help
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  4. #4
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    Re: Prove K^2 mod 4 = 1 ?

    Basically 4m + 1 where m is an integer implies congruence to 1 (MOD 4) and you're done.
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  5. #5
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    Re: Prove K^2 mod 4 = 1 ?

    Thank you very very much for your help chiro, but I'm not understanding what your saying.

    Can you show me what the entire proof would be (for Christmas?).
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  6. #6
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    Re: Prove K^2 mod 4 = 1 ?

    Let k = 2n + 1 where n is an integer (in Z).

    Then k^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1 for some integer m (m in Z).

    By the definition of congruence something is congruent z = r (MOD q) iff z = pq + r where 0 <= r < q for some integers p, r, q and z.

    By letting z = k^2, p = m, r = 1, and q = 4 and noting 0 <= 1 < 4 and all variables are integers, we get our condition that

    k^2 = (1 MOD 4)
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