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Math Help - Division proof

  1. #1
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    Division proof

    Prove that if a is an odd integer, then 24 | a(a^2-1).

    Proof. from a previous in my homework, I know that the square of an odd integer can be written in the form of 8k + 1.

    Now, let a = 2w + 1 and a^2 = 8k + 1, for some integers w and k.

    Then a(a^2-1) = (2w+1)[(8k+1)^2-1] = 128k^2w+32kw+64k^2+16k

    But now, how do I factor out 24?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Prove that if a is an odd integer, then 24 | a(a^2-1).

    Proof. from a previous in my homework, I know that the square of an odd integer can be written in the form of 8k + 1.

    Now, let a = 2w + 1 and a^2 = 8k + 1, for some integers w and k.

    Then a(a^2-1) = (2w+1)[(8k+1)^2-1] = 128k^2w+32kw+64k^2+16k

    But now, how do I factor out 24?
    Since a is odd it means a=2b+1 and so a(a^2-1) = a(a-1)(a+1) = (2b+1)(2b)(2b+2) = 4b(b+1)(2b+1). Thus, 24 divides 4b(b+1)(2b+1) is equivalent to saying 6 divides b(b+1)(2b+1). Now since 6=2\cdot 3 it means 6|b(b+1)(2b+1) if and only if 2|b(b+1)(2b+1) \mbox{ and }3|b(b+1)(2b+1) because \gcd (2,3) = 1. Can you finish now?
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  3. #3
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    I prefer to do it this way. Notice that a-1 and a+1 are consecutive even integers since a is odd. Hence the product (a-1)(a+1)=a^2-1 is divisible by 8. (NB: the product of two consecutive even intgers 2k and 2k+2 is 4k(k+1). One of k and k+1 is even and so 4k(k+1) must be divisible by 8.)

    Also a-1,a,a+1 are three consecutive integers and so one of them must be a multiple of 3.

    Putting the two together, we have that the product (a-1)a(a+1)=a(a^2-1) is divisible by both 8 and 3. Now the proof is almost finished. Iíll let you finish it yourself.
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    So since gcd(3,8) = 1, (3)(8) = 24 | a(a^2-1)
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    So since gcd(3,8) = 1, (3)(8) = 24 | a(a^2-1)
    You can do it that way too. Let me state the result you are using: let a and b be positive integers such that \gcd(a,b)=1 then if a|c and b|c then ab|c. Thus, since 24 = (8)(3) and \gcd(3,8)=1 it means if you can prove that 3|a(a^2-1) and 8|a(a^2-1) then 24|a(a^2-1).
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