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Induction seems to not work?

  1. Jun 11, 2008 #1


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    1. The problem statement, all variables and given/known data

    Prove by mathematical induction,or otherwise,that [itex]23^{2n}+31^{2n}+46[/itex] is divisible by 48 for all [itex]n \geq 0[/itex]

    2. Relevant equations

    3. The attempt at a solution

    Assume true for n=N


    [tex] \times (23^2 + 31^2)[/tex]

    [tex]23^{2N+2}+31^{2N+2}+46(23^2 + 31^2)+23^2(31^{2N})+31^2(23^{2N})=48A(23^2 + 31^2)[/tex]

    [tex]23^{2N+2}+31^{2N+2}+68540+31^{2N}(528+1)+23^{2N}(960+1)=48A(23^2 + 31^2)[/tex]

    Simplifying gives me what I need, i.e. things divible by 48, but 68540 is not...any suggestions?
  2. jcsd
  3. Jun 11, 2008 #2


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    Everything is not supposed to be divisible by 48. You can drop the things that are, of course. Now you just want to show 68540+31^(2N)+23^(2N) gives a remainder of 46 when divided by 48. BTW, it's easier without induction. Just observe 23^2 and 31^2 are both equal to 1 mod 48.
  4. Jun 11, 2008 #3
    Code (Text):

    what's this doing there? if this is a part of the first equation ...

    at n+1:
    23^(2n)*23^2 + 31^(2n)*31^2+46

    now trying writing this in orginal form:

    23^(2n)*528 + 23^(2n) + 31^(2n)*31^2+46

    do same for 31 ..
  5. Jun 11, 2008 #4
    since you already got the answer.. here's what I did:

    23^(2n)*528 + 23^(2n) + 31^(2n)*960 + 31 ^(2n) +46

    = {23^(2n) + 31 ^(2n) +46} + {23^(2n)*528 + 31^(2n)*960}

    everything divisible by 48
  6. Jun 11, 2008 #5


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    Well I multiplied what I assumed to be true by [itex]23^2+31^2[/itex] so could get [itex]23^{2N+2}+31^{2N+2}[/itex]

    wouldn't I need to show that everything is divisible by 48 for it to be true,hence why I can't show 68540 is divisible by 48.

    I don't get how I am supposed to show it gives a remaineder of 48, when in the inductive hypothesis, I assumed that everything was together was divisible by 48.

    Also,I don't understand what 1 mod 48 means. If it has to do with modular arithmetic,then I definately don't know what you are talking about since I never learnt that.
  7. Jun 11, 2008 #6

    Over there, from induction you assumed that {23^(2n) + 31 ^(2n) +46} is divisible by 48

    at n+1 .. you get {23^(2n) + 31 ^(2n) +46} + {23^(2n)*528 + 31^(2n)*960}

    divide this by 48..
    {23^(2n) + 31 ^(2n) +46} is divisible
    23^(2n)*528 is divisible
    31^(2n)*960 is divisible...

    so everything is proved
  8. Jun 11, 2008 #7


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    So I am supposed to just substitute n=N+1 instead of building the (N+1)th term?
  9. Jun 11, 2008 #8
    yep, isn't this is what you do for induction?

    first assume it is true for n
    and then substitute n by n+1 .. and prove it is true
  10. Jun 11, 2008 #9


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    You really already have it, you just don't seem to see it. According to your inductive hypothesis, if you divide 23^(2N)+31^(2N) by 48, what is the remainder? Now look at the N+1 case. Assuming the hypothesis what's the remainder of 23^(2N+2)+31^(2N+2) divided by 48?
  11. Jun 12, 2008 #10


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    He didn't say "remainder of 48", he said remainder of 46- so that added to the other remainders, they add up to 48.

    232= 529= 11(48)+ 1, 312= 961= 20(48)+ 1
    So each has a remainder of one when divided by 48.

    Also (x+ 1)n= a string of things times powers of x + 1 and so has a remainder of 1 when divided by x. Both 232n and 312n have a remainder of 1 when divided by 48 for any n.
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