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Deductive proofs

  1. Sep 6, 2006 #1
    Here is the question. I have to prove it.

    Prove that the square of an odd integer is always of the form 8k+1, which k is an integer.

    Now I do not know how to start it. But this is what I came up with.

    odd integer= 2k+1

    therefore the square of an odd integer (2k+1)^2

    i have used inductive reasoning to prove that is statement is correct
    if k=1 then the expression becomes 9 (3^2)
    if k =3 then the epression becomes 25 (5^2)
    if k=6 then the expression becomes 49 (7^2)

    now how to prove it with deductive reasonings.

    I am not sure how to start this one.

    Can you please give me a hint:rolleyes:
  2. jcsd
  3. Sep 6, 2006 #2


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    That is not induction, that is just trying a few examples and it shows almost nothing. (though it may help for thinking about it) Mathematical induction is rigorous and deductive--have you been exposed to it?

    Anyway, it may be easier in this case not to use induction. What happens if you expand the square? How can you rewrite that so that you know part of it is divisible by 8, and the other part is 1?
  4. Sep 6, 2006 #3
    thanks for the help

    i worked out the problem to get

    4 k (k+1) + 1

    therefore the one term is divisible by 4, and 2 (since k and k+1 are consecutive making one of them even)

    now i have a remainder of 1, does that matter. why did you tell me to have one part of the equation 1.
  5. Sep 6, 2006 #4


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    The problem states,
    "Prove that the square of an odd integer is always of the form 8k+1, where k is an integer."
    In what you've written, can you identify the 8k and the 1? (the "k" will be a different number from what you have for k)
  6. Sep 7, 2006 #5


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    You are confusing things by using the same letter, k, in the statement "the square of an odd number is 8k+ 1" and "the odd number is 2k+1".
    Also you then calculate for various values of k in the 8k+ 1 form. You want to look the other way: what happens for different k in (2k+1)2?

    Rewrite the statement of theorem as "the square of any odd number is 8n+1".

    12= 1= 8(0)+ 1 so it is true.

    Assume (2k+1)2= 4k2+ 4k+ 1= 8n+ 1.

    Now look at k+1 instead of k: the next odd number is 2(k+1)+ 1= 2k+ 3 and (2k+ 3)2= 4k2+ 12k+ 9. Can you rewrite that as (4k2+ 4k+ 1) plus something?
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