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Induction question

  1. Feb 2, 2006 #1
    Prove by induction that [tex]2^{2^n}+1[/tex] always ends in 7 for all n > 1 (true for n = 2).

    I couldn't figure out anything to do with the last digit being 7, so I looked the case that [tex]2^{2^n}[/tex] ends in 6 for all n > 1, which is also true for n = 2.

    Suppose it's true for n = k:

    [tex]2^{2^{k+1}}=2^{2^k\cdot 2}=(2^{2^k})^2[/tex]

    Now, since by assumption [tex]2^{2^k}[/tex] ends in 6, its 2nd power also ends in 6 since 6x6=36. Hence [tex]2^{2^n}+1[/tex] ends in 7 for all n > 1.

    Correct or not?

    Thanks in advance,

    - Kamataat
  2. jcsd
  3. Feb 2, 2006 #2


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    Yes, that's perfectly good. Obviously n+ 1 will end in 7 if and only if n ends in 6.

    You could, just as easily, say "Suppose [itex]2^{2^k}+1[/itex] ends in 7 for some k. Then [itex]2^{2^k}[/itex] ends in 6. Therefore, [itex]2^{2^{k+1}}= (2^{2^k})^2[/itex] also ends in 6 and so [itex]2^{2^{k+1}}+ 1[/itex] ends in 7".
    Last edited: Feb 2, 2006
  4. Feb 2, 2006 #3
    I think what you've done is correct. Here is how I would set it out. I'm not sure if it's correct though.

    Base case, assume true for n = so that [itex]2^{2^k } \equiv 6\left( {\bmod 10} \right)[/itex]

    Using the hypothesis:

    2^{2^{\left( {k + 1} \right)} } \equiv 2^{2\left( {2^k } \right)} \equiv \left( {2^{2^k } } \right)^2 \equiv \left( 6 \right)^2 \equiv 6\left( {\bmod 10} \right)

    I think the result follows from that.
  5. Feb 2, 2006 #4


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    Oops. Stumbled into the wrong thread. Though it was about labour & delivery. Please continue...
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