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

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

    Putting the two together, we have that the product $\displaystyle (a-1)a(a+1)=a(a^2-1)$ 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 $\displaystyle a$ and $\displaystyle b$ be positive integers such that $\displaystyle \gcd(a,b)=1$ then if $\displaystyle a|c$ and $\displaystyle b|c$ then $\displaystyle ab|c$. Thus, since $\displaystyle 24 = (8)(3)$ and $\displaystyle \gcd(3,8)=1$ it means if you can prove that $\displaystyle 3|a(a^2-1)$ and $\displaystyle 8|a(a^2-1)$ then $\displaystyle 24|a(a^2-1)$.
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