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More Proofs: Prove that if n is an odd positive int., then n^2 = 1(mod 8)

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that if n is an odd positive integer, then [itex]n^2\,\equiv\,1\,\left(mod\,8\right)[/itex].

    2. Relevant equations



    if and only if


    3. The attempt at a solution

    Using the theorem above:



    Then I have this:


    What do I do next to show this for odd positive n?
  2. jcsd
  3. Apr 5, 2007 #2


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    This is hardly "beyond calculus". Anyways, you can represent odd numbers with 2k + 1, where k is a natural number (or zero). Then use eg. proof by induction.
  4. Apr 5, 2007 #3


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    You need to prove that [itex] (2k+1)^2=8p+1 [/itex] for some "k" and "p" in [itex] \mathbb{Z} [/itex].
  5. Apr 5, 2007 #4
    How would I go about proving that?



    Now what?
  6. Apr 5, 2007 #5


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    If n is an odd number, then n can be expressed by (2k + 1) where k is any whole number. Right?
    So, we have:
    n2 = (2k + 1)2 = 4k2 + 4k + 1
    = 4 (k2 + k) + 1 = 4k(k + 1) + 1
    What can you say about the product of 2 successive integers? Hint: Is that divisible by 2?
    So, is the product 4k(k + 1) divisible by 8?
    Can you go from here? :)
  7. Apr 5, 2007 #6


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    p is any integer, so k^2+k=2p is equivalent to (ie, exactly the same question as) whether k^2+k is even. Well, is it?
  8. Apr 5, 2007 #7
    [itex]k^2\,+\,k[/itex] is always even. If k were odd, adding an odd integer to another odd one produces an even integer. If k were even, then the whole expression is even as well because adding two even integers always results in an even integer.
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