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Thread: Prove congruence for powers of 2

  1. #1
    MHF Contributor alexmahone's Avatar
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    Prove congruence for powers of 2

    Let $n=2^k$ where $k\ge 3$. Let $a$ be any odd natural number.

    Prove that $a^{n/4}\equiv 1\pmod{n}$.

    My attempt:

    $\phi(n)=n/2$

    So, by Euler's formula, $a^{n/2}\equiv 1\pmod{n}$.

    I don't know how to proceed.
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  2. #2
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    Re: Prove congruence for powers of 2

    Quote Originally Posted by alexmahone View Post
    Let $n=2^k$ where $k\ge 3$. Let $a$ be any odd natural number.

    Prove that $a^{n/4}\equiv 1\pmod{n}$.

    My attempt:

    $\phi(n)=n/2$

    So, by Euler's formula, $a^{n/2}\equiv 1\pmod{n}$.

    I don't know how to proceed.
    if $a \equiv 1 \pmod{n}$

    what is $a^2 \pmod{n}$ ?
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  3. #3
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    Re: Prove congruence for powers of 2

    a^{2^{k-1}}-1=\left(a^{2^{k-2}}-1\right)\left(a^{2^{k-2}}+1\right)

    use induction on k
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Prove congruence for powers of 2

    Quote Originally Posted by Idea View Post
    a^{2^{k-1}}-1=\left(a^{2^{k-2}}-1\right)\left(a^{2^{k-2}}+1\right)

    use induction on k
    Thanks!
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