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Prove 5|(3^(3n+1)+2^(n+1)) for every positive integer n.

  1. Mar 18, 2014 #1
    "Prove ##5|(3^{3n+1}+2^{n+1})## for every positive integer ##n##." (Exercise 11.8 from Mathematical Proofs: A Transition to Advanced Mathematics 3rd edition by Chartrand, Polimeni & Zhang; pg. 282)

    I'm having difficulty solving this exercise. My first thought was to use induction but I get stuck with an expression that I can't seem to manipulate into the required form. The next thing I tried was considering cases (when ##n## is even and when ##n## is odd) but once again I get stuck.

    Can anyone suggest another method of proof?

    I feel that induction should work out for me despite my difficulties. As such, I will explain what I've done so far. If you would like to see work I describe but do not include below let me know and I'll include it.


    First I showed that this expression holds for the case ##n=1##. I assumed this holds for some arbitrary positive integer ##k##. Now I wish to show that ##5|(3^{3(k+1)+1}+2^{(k+1)+1})##.

    From here I tried working with the expression ##3^{3(k+1)+1}+2^{(k+1)+1}##:

    ##3^{3(k+1)+1}+2^{(k+1)+1}=3^{3k+3+1}+2^{k+1+1}=3^{3k+4}+2^{k+2}=3^{3k} \cdot 3^4 + 2^{k} \cdot 2^{2}=81\cdot3^{3k}+4\cdot2^{k}##

    This is where I get stuck - I don't know how to massage this equation to show that 5 divides it.
     
  2. jcsd
  3. Mar 18, 2014 #2

    Dick

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    Write that as ##3^{3(k+1)+1}+2^{(k+1)+1}=27*3^{3k+1}+2^{k+1}##. Note 27=25+2.
     
  4. Mar 18, 2014 #3
    Thanks so much for the hint!

    So we can write:

    ##3^{3(k+1)+1}+2^{(k+1)+1}##
    ##= 27 \cdot 3^{3k+1}+2\cdot 2^{k+1}##
    ##=(25+2)\cdot 3^{3k+1}+2 \cdot 2^{k+1}##
    ##=25 \cdot 3^{3k+1} + 2 \cdot 3^{3k+1} + 2\cdot 2^{k+1}##
    ##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##

    By assumption we know that ##5|(3^{3k+1} + 2^{k+1})## so we can write ##(3^{3k+1} + 2^{k+1})=5x## for some integer x.

    Then we have
    ##= 25 \cdot 3^{3k+1} + 2 \cdot (3^{3k+1} + 2^{k+1})##
    ##=5(5\cdot3^{3k+1})+5\cdot2x##
    ##=5(5\cdot3^{3k+1}+2x)##

    Since ##5\cdot3^{3k+1}+2x## is an integer, we conclude that ##5|3^{3(k+1)+1}+2^{(k+1)+1}##.
     
    Last edited: Mar 18, 2014
  5. Mar 18, 2014 #4

    Dick

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    Right. You definitely want to keep ##3^{3k+1} + 2^{k+1}## in the answer somehow because you know it's divisible by 5. And splitting the factor up becomes pretty obvious once you've seen the trick.
     
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