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Math Help - Pseudoprimes

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    Post Pseudoprimes

    Prove that n is an odd pseudoprime number, then m=(2^n) - 1 is an odd pseudoprime number.
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    Quote Originally Posted by jboite View Post
    Prove that n is an odd pseudoprime number, then m=(2^n) - 1 is an odd pseudoprime number.
    Since n is an odd pseudoprime it means 2^n\equiv 2(\bmod n), hence 2^n - 2 = kn for some k\in \mathbb{Z}.

    Now m=2^n - 1 is not prime since n is not prime, so it sufficies to show 2^m\equiv 2(\bmod m). Note that 2^{m-1} = 2^{2^n - 2} = 2^{kn}. Therefore, 2^{m-1} - 1 is divisible by 2^n - 1=m since n|(m-1). Thus, 2^{m-1} \equiv 1(\bmod m) \implies 2^m \equiv 2(\bmod m).
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