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Thread: Proofs-Divisibility of Integers

  1. #1
    Junior Member utopiaNow's Avatar
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    Proofs-Divisibility of Integers

    Let $\displaystyle n \in \mathbb{Z}$. Prove that $\displaystyle 2\ |\ (n^4 - 3) $ if and only if $\displaystyle 4\ |\ (n^2 + 3) $.

    Attempt at Proof:
    First I just dealt with the forward implication: if $\displaystyle 2 \ |\ (n^4 - 3) $ then $\displaystyle 4 \ |\ (n^2 + 3) $. To prove this I took the contrapositive of this statement which is: If $\displaystyle 4 \not\vert\ (n^2 + 3) $, then $\displaystyle 2 \not\vert\ (n^4 - 3) $.
    $\displaystyle 4 \not\vert\ (n^2 + 3) $ can be broken down into 3 cases, the case I'm having trouble with is: let $\displaystyle (n^2 + 3) = 4q + 2$, then $\displaystyle n^2 = 4q - 1 $ and then $\displaystyle n^4 - 3 = (4q - 1)^2 - 3 = 16q^2 - 8q - 2 = 2(8q^2 -4q - 1) $ which is an even number and so $\displaystyle 2 \ |\ (n^4 - 3) $, but that's not what I wanted for the contrapositive.

    What am I doing wrong?
    Thanks in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by utopiaNow View Post
    Let $\displaystyle n \in \mathbb{Z}$. Prove that $\displaystyle 2\ |\ (n^4 - 3) $ if and only if $\displaystyle 4\ |\ (n^2 + 3) $.

    Attempt at Proof:
    First I just dealt with the forward implication: if $\displaystyle 2 \ |\ (n^4 - 3) $ then $\displaystyle 4 \ |\ (n^2 + 3) $. To prove this I took the contrapositive of this statement which is: If $\displaystyle 4 \not\vert\ (n^2 + 3) $, then $\displaystyle 2 \not\vert\ (n^4 - 3) $.
    $\displaystyle 4 \not\vert\ (n^2 + 3) $ can be broken down into 3 cases, the case I'm having trouble with is: let $\displaystyle (n^2 + 3) = 4q + 2$, then $\displaystyle n^2 = 4q - 1 $ and then $\displaystyle n^4 - 3 = (4q - 1)^2 - 3 = 16q^2 - 8q - 2 = 2(8q^2 -4q - 1) $ which is an even number and so $\displaystyle 2 \ |\ (n^4 - 3) $, but that's not what I wanted for the contrapositive.

    What am I doing wrong?
    Thanks in advance.
    $\displaystyle 2\ |\ (n^4 - 3) $ implies that $\displaystyle n^4$ is odd which implies $\displaystyle n$ is odd.

    If $\displaystyle n$ is odd there exists a $\displaystyle k \in \mathbb{Z}$ such that $\displaystyle n=2k+1$

    Then:

    $\displaystyle n^2+1=(2k+1)^2+3=4k^2+4k+4$

    The other direction of the proof proceeds by observing that $\displaystyle 4\ |\ (n^2 + 3) $ implies $\displaystyle n$ odd ...

    CB
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