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Thread: Possible mathematical induction problem

  1. #1
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    Possible mathematical induction problem

    Course: Foundations of Higher MAth

    Prove that $\displaystyle 24|(5^{2n} -1)$ for every positive integer n.

    This is a question from my final exam today.

    P(n): $\displaystyle 24|(5^{2n}-1)$
    P(1): $\displaystyle 24|(5^2 -1)$ is a true statement.

    Assume P(k) is true. Then $\displaystyle 5^{2k}-1 = 24a$ for some integer a.
    Then $\displaystyle 5^{2k}= 24a + 1$

    P(k+1): $\displaystyle 24|(5^{2(k+1)}-1)$

    $\displaystyle 5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
    $\displaystyle =(24a + 1)*25 -1$
    $\displaystyle =(24a)(25)+25 -1$
    $\displaystyle =(24a)(25)+24$
    $\displaystyle =24(25a+1)$
    $\displaystyle =24b$

    Therefore, $\displaystyle 24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
    None of my friends used this method though. Is this a correct way to do it?
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  2. #2
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    Re: Possible mathematical induction problem

    Quote Originally Posted by MadSoulz View Post
    Course: Foundations of Higher MAth
    Prove that $\displaystyle 24|(5^{2n} -1)$ for every positive integer n.
    This is a question from my final exam today.
    P(n): $\displaystyle 24|(5^{2n}-1)$
    P(1): $\displaystyle 24|(5^2 -1)$ is a true statement.
    Assume P(k) is true. Then $\displaystyle 5^{2k}-1 = 24a$ for some integer a.
    Then $\displaystyle 5^{2k}= 24a + 1$
    P(k+1): $\displaystyle 24|(5^{2(k+1)}-1)$
    $\displaystyle 5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
    $\displaystyle =(24a + 1)*25 -1$
    $\displaystyle =(24a)(25)+25 -1$
    $\displaystyle =(24a)(25)+24$
    $\displaystyle =24(25a+1)$
    $\displaystyle =24b$
    Therefore, $\displaystyle 24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
    None of my friends used this method though. Is this a correct way to do it?
    A strict grader may like to have seen more grouping symbols.
    However, the argument is correct.
    Thanks from MadSoulz
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