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Prove, using induction

  1. Mar 23, 2005 #1
    I was asked to prove, using induction, that 34n+2 + 26n+3 is divisible by 17. I tried to do it, but I couldn't get anywhere. Can someone give me a push in the right direction?

    Here's my attempt:
    f(k) = 34k+2 + 26k+3
    f(k+1) = 34k+6 + 26k+9
    And now, all I have to do is prove that f(k+1) - f(k) = 17m, but I couldn't do it.

    I don't really see why induction is necessary anyway. Here's my induction-free attempt:

    [tex]3^{4n+2} + 2^{6n+3} = 9^{2n+1} + 8^{2n+1} = (9+8) \sum^{2k+1}_{n=1} 9^{2k+1-n} \; 8^{n-1} = 17m[/tex]
  2. jcsd
  3. Mar 23, 2005 #2


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    Your proof is incorrect - the factoring is wrong.

    Proof of theorem: f(k+1)=81*34k+2+64*26k+3
    =17*34k+2+64*f(k), where both terms in the sum are divisible by 17. You need also f(0)=17 to start with.
  4. Mar 24, 2005 #3

    By the way, did you mean that my induction-free proof was wrong?
  5. Mar 24, 2005 #4


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    Yes: 92n+1+82n+1 does not factor as you wrote it. You confused it with an expression with a minus sign.
  6. Mar 24, 2005 #5
    Ah yes, it's:
    a^n + b^n = (a+b)[a^(n-1) - a^(n-2)b + ... + b^(n-1)], correct?
  7. Mar 24, 2005 #6


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    You mean like a^2+ b^2= (a+ b)(a-b )???

    Oops, I guess it's not correct.
  8. Mar 24, 2005 #7


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    Nope,it the REALS,you can't factor [tex] a^{n}+b^{n} [/tex]...

  9. Mar 24, 2005 #8
    Well, if n is odd you can, and his newest attempt at a factorization is correct. This makes his original proof right, with a small modification.

    [tex]n \in \mathbb{N} \Longrightarrow 3^{4n+2} + 2^{6n+3} = 9^{2n+1} + 8^{2n+1} = 17\sum_{k=0}^{2n}(-1)^{k} \ 9^k \ 8^{2n-k}[/tex]
    Last edited: Mar 24, 2005
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