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Math Help - Induction

  1. #1
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    Induction

    I have the question prove by induction that 5^2n-4^n is divisible by 7.

    In proving p(n+1) implies P(n) I have got:

    (5^2n-4^n)+ 21(5^2n-4^n) is divisible by 7 by inductive hypothesis and since 21 is divisible by 7

    Just wondering if this argument holds or if anyone has any ideas?
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    Quote Originally Posted by iwish123 View Post
    I have the question prove by induction that 5^2n-4^n is divisible by 7.

    In proving p(n+1) implies P(n) I have got:

    (5^2n-4^n)+ 21(5^2n-4^n) is divisible by 7 by inductive hypothesis and since 21 is divisible by 7

    Just wondering if this argument holds or if anyone has any ideas?
    "p(n+1) implies P(n)" is backwards, should read "p(n) implies p(n+1)"

    This is how my inductive step went:

    5^{2(n+1)}-4^{n+1}=25\cdot5^{2n}-4\cdot4^n\equiv4\cdot5^{2n}-4\cdot4^n\equiv4(5^{2n}-4^n)\equiv0\ \text{(mod }7\text{)}\
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    Quote Originally Posted by iwish123 View Post
    I have the question prove by induction that 5^2n-4^n is divisible by 7.

    In proving p(n+1) implies P(n) I have got:

    (5^2n-4^n)+ 21(5^2n-4^n) is divisible by 7 by inductive hypothesis and since 21 is divisible by 7

    Just wondering if this argument holds or if anyone has any ideas?
    P(k)

    5^{2k}-4^k is divisible by 7

    P(k+1)

    5^{2(k+1)}-4^{k+1} is also divisible by 7.

    You can attempt to prove that 5^{2k}-4^k being divisible by 7 causes 5^{2k+2}-4^{k+1} to be divisible by 7.

    This then means that if your formula is true for some n=k, it's automatically true for the next n=k+1.

    Hence, if it's then true for n=1, it's true for n=2, 3, 4, 5, 6, 7, 8, to infinity.

    5^25^{2k}-(4)4^k=(25)5^{2k}-(4)4^k

    We write this "in terms of" 5^{2k}-4^k to get

    (21+4)5^{2k}-(4)4^k=(21)5^{2k}+4\left(5^{2k}-4^k\right)

    Hence, if 5^{2k}-4^k really is divisible by 7,

    then, since 21 is (7)3

    5^{2(k+1)}-4^{k+1} will also be divisible by 7.
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