use induction to prove k^(2n)≡1(mod 2^(n+2)) for any odd integer k?

use induction to prove k^(2n)≡1(mod 2^(n+2)) for any odd integer k

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  • Anonymous
    7 years ago
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    Base step: Let n = 1. Then, we have...

    k² ≡ 1 mod(2³)

    k² ≡ 1 mod(8)

    This implies:

    k² = 1 + 8m for every integer m

    Suppose k = 2p + 1 for every p integer. Then, we have...

    (2p + 1)² = 1 + 8m where m is the positive integer

    4p² + 4p + 1 = 1 + 8m

    4p² + 4p = 8m

    4p² + 4p - 8m = 0

    4(p² + p - 2m) = 0

    So for any p integer, the statement holds...

    This goes same for inductive step [Let n = q + 1. Then, follow the similar method]

    Good luck!

    Source(s): τ
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  • 3 years ago

    by way of fact of certainty the effective integers initiate up at n = a million, that's our base case. With n = a million, we see that the LHS is two^2 = 4 on an analogous time by way of fact the RHS is two(a million)(a million + a million)(2 + a million)/3 = 4. This exhibits that the backside case is authentic. The inductive step is to coach that P(n + a million) holds if P(n) holds. it extremely is: 2^2 + 4^2 + 6^2 + ... + (2n)^2 + [2(n + a million)]^2 = 2(n + a million)[(n + a million) + a million][2(n + a million) + a million]/3, given 2^2 + 4^2 + 6^2 + ... + (2n)^2 = 2n(n + a million)(2n + a million)/3. Then, utilising this: 2^2 + 4^2 + 6^2 + ... + (2n)^2 + [2(n + a million)]^2 = [2^2 + 4^2 + 6^2 + ... + (2n)^2] + [2(n + a million)]^2 = 2n(n + a million)(2n + a million)/3 + [2(n + a million)]^2, via the inductive hypothesis = 2n(n + a million)(2n + a million)/3 + 4(n + a million)^2 = [2n(n + a million)(2n + a million) + 12(n + a million)^2]/3 = 2(n + a million)[n(2n + a million) + 6(n + a million)]/3 = 2(n + a million)(2n^2 + 7n + 6)/3 = 2(n + a million)(n + 2)(2n + 3)/3, via factoring = 2(n + a million)[(n + a million) + a million][2(n + a million) + a million]/3. by way of fact of certainty the backside case and the inductive step carry, we are able to end the info. i desire this helps!

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