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Math Help - Stuck on a proof

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    33

    Stuck on a proof

    I'm stuck on the final step in these two proofs:
    1) For every natural number N that leaves remainder of 1 when divided by 5, there's a n^2 that also leaves remainder of 1 when divided by 5

    n = 5k + 1 where k, an integer, is the quotient:
    Then n^2 = (5k + 1)^2
    n^2 = 25k^2 + 10k + 1
    = 5(5k^2 + 2k) + 1

    Stuck here...

    let j be 5k^2 + 2k
    if k is an integer, then 5k^2 + 2k is also an integer
    n^2 = 5j + 1, where j is an integer quotient


    And the very similar problem...


    2) For every natural number N that leaves remainder of 4 when divided by 5, there's a n^2 that leaves remainder of 1 when divided by 5

    n = 5j + 4 where j, an integer, is the quotiet:
    Then n^2 = (5j + 4)^2
    n^2 = 25j^2 + 40j + 16
    = 25j^2 + 40j + 15 + 1
    = 5(5j^2 + 8j + 3) + 1
    = 5(j + 1)(5j + 3) + 1


    Stuck here...




    Thanks for any help.
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  2. #2
    Member
    Joined
    Mar 2009
    Posts
    90
    hi selena,

    you've proved it! There's nothing more to do... you might want to think about how you present the proof. The implication is:
    <br />
(\forall n)(\exists j, k)(n = 5j + 1) \Rightarrow (n^2 = 5k + 1); n, j, k \in \mathbb{N}<br />

    and then you use your proof to show:
    <br />
n^2 =  5(5j^2 + 2k) + 1<br />

    and notice that k = 5j^2 + 2j \Rightarrow k \in \mathbb{N}. You normally don't have to show that any two integers add or multiply to give another integer.
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