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Math Help - Proofs-Divisibility of Integers

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

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

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

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

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

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

    Then:

    n^2+1=(2k+1)^2+3=4k^2+4k+4

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

    CB
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