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Thread: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

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    Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Course: Foundations of Higher Mathematics

    Prove that $\displaystyle 2|(n^4 - 3)$ if and only if $\displaystyle 4|(n^2+3)$

    I feel confident about this one. It's only the first part of the biconditional. Here's my attempt.

    Assume that $\displaystyle 2|(n^4-3)$ , i.e. $\displaystyle n^4-3=2a$, for some integer a.

    Then, $\displaystyle n^4=2a+3$
    $\displaystyle =2(a+1) +1$
    $\displaystyle =2b+1$.
    Since $\displaystyle a+1$ is an integer, b is an integer also.

    Therefore $\displaystyle n^4$ is odd, which means that $\displaystyle n$ is odd, i.e. $\displaystyle n=2c+1$, for some integer c.

    Then, $\displaystyle n^2+3=(2c+1)^2+3$
    $\displaystyle =4c^2+4c+4$
    $\displaystyle =4(c^2+c+1)=4d$, for some integer d

    Therefore $\displaystyle n^2+3=4d$, hence $\displaystyle 4|(n^2+3)$
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    Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Nicely done.
    Thanks from MadSoulz
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    Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Quote Originally Posted by MadSoulz View Post
    Prove that $\displaystyle 2|(n^4 - 3)$ if and only if $\displaystyle 4|(n^2+3)$
    My take, which I don't know how to finish:
    From the first relation: $\displaystyle n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $\displaystyle n \equiv 1 \text{ (mod 2)}$

    And from the second: $\displaystyle n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $\displaystyle n \equiv {1, 3} \text{ (mod 4)}$

    How do I finish? Is it the Chinese remainder theorem?

    -Dan
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    Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Quote Originally Posted by topsquark View Post
    My take, which I don't know how to finish:
    From the first relation: $\displaystyle n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $\displaystyle n \equiv 1 \text{ (mod 2)}$

    And from the second: $\displaystyle n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $\displaystyle n \equiv {1, 3} \text{ (mod 4)}$

    How do I finish? Is it the Chinese remainder theorem?

    -Dan
    I'm not too familiar with modulo just yet. And I've never heard of that theorem before.
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    Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Quote Originally Posted by topsquark View Post
    From the first relation: $\displaystyle n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $\displaystyle n \equiv 1 \text{ (mod 2)}$

    And from the second: $\displaystyle n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $\displaystyle n \equiv {1, 3} \text{ (mod 4)}$
    I am not sure which fact stated purely in terms of ≡ to use, but it is clear that n ≡ 1, 3 (mod 4) iff n is odd.
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    Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

    Quote Originally Posted by emakarov View Post
    I am not sure which fact stated purely in terms of ≡ to use, but it is clear that n ≡ 1, 3 (mod 4) iff n is odd.
    I can't believe I missed that. Thanks!

    -Dan
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