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Math Help - Number theory integer proofs

  1. #1
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    Unhappy Number theory integer proofs

    Some number theory problems:
    1. Show that every integer of the form 6k+5 must also be of the form 3k+2
    2. Show that the sqaure of any odd integer is of the form 8k+1
    3. Prove that the cube of any integer can be written as one of 9k-1,9k or 9k+1
    4. Prove that the sum of the squares of two odd integers cannot be a perfect square.
    5. Prove that the difference of two consectutive cubes is odd.

    Any help would be gratefully accepted.

    Cheers
    Cabouli
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  2. #2
    MHF Contributor chisigma's Avatar
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    Show that every integer of the form 6k+5 must also be of the form 3k+2…

    Setting n=6k+5 we have immediately…

    n=5 \mod 6 = 2 \mod 3

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  3. #3
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    Hello, Cabouli!

    Here are a few of them . . .


    2. Show that the square of any odd integer is of the form 8k+1
    An odd integer has the form: . N = 2a+1\,\text{ for some integer }a.

    Its square is: . N^2 \:=\:(2a+1)^2 \:=\:4a^2 + 4a + 1

    We have: . N^2 \:=\:4\underbrace{a(a+1)} + 1

    Note that a and a+1 are consecutive integers.
    So one of them is even and the other is odd.
    Hence: . a(a+1) is an even integer, 2k,

    And we have: . N^2 \;=\;4(2k)+1 \;=\;8k+1




    4. Prove that the sum of the squares of two odd integers cannot be a perfect square.
    If an integer is even, 2a,
    . . its square is: . 4a^2, a multiple of 4.

    If an integer is odd, 2a+1,
    . . its square is: . 4a^2 + 4a + 1 \:=\:4a(a+1) + 1, one more than a multiple of 4.

    Hence, the square of an integer has the form 4k\,\text{ or }\,4k+1


    The sum of the squares of two odd integers is:
    . . (2a+1)^2 + (2b+1)^2 \:=\:4a^2 + 4a + 1 + 4b^2 + 4b+1 \:=\: 4(a^2+b^2+a+b) + 2

    This is two more than a multiple of 4 . . . It cannot be a square.




    5. Prove that the difference of two consectutive cubes is odd.
    The two consecutive cubes are: . a^3\,\text{ and}\,(a+1)^3

    Their difference is: . (a+1)^3-a^3 \:=\:a^3 + 3a^2 + 3a + 1 - a^3 \:=\:3a^2 + 3a + 1

    \text{We have: }\;3\underbrace{a(a+1)} + 1
    The product of two consecutive integers is even, 2k

    \text{So we have: }\;3(2k) + 1 \:=\:\underbrace{6k}_{\text{even}} + 1 \:=\: \text{odd}

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  4. #4
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    Smile Thank you

    Thank you for all your help.

    Cheers
    Cabouli
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