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Math Help - Number theory powers of numbers proof

  1. #1
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    Number theory powers of numbers proof

    Find all positive integers k and l such that 3^{k} - 2^{l} = 1

    This is the last part of a question and I have previously proved that 2^{l} + 1 is divisible by 27 iff it is also divisible by 19 and did this by examining the sequence of powers of 2 reduced modulo 27 and the sequence of powers of 2 reduced modulo 19.
    I'm assuming this somehow relates to this proof but not sure how to start?
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  2. #2
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    Well if you refer to the tables you drew out from parts i,ii...

    you will notice some connection between them.. write 27 as 3^3 so

    Also, look at how 3^k = 1+ 2^l

    we can see when k=1, l=1 it gives us a solution..

    Such a small world!
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  3. #3
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     k=l=1 is a solution and suppose  l \geq 2 so 2^l \equiv 0 \bmod{4} . Take modulo 4 , we have :

     (-1)^k \equiv 1 \bmod{4} which forces  k to be even .

    Let  k = 2x and consider the factorization :

     3^{2x} - 1 = 2^l = (3^x - 1)(3^x + 1 ) . We have to set both of  3^x - 1 and  3^x + 1 to be the power of two because of the other side . Call them  2^s and  2^t respectively , so we have  2^t - 2^s = 2 , the only solution to this easy equation is  t=2 ~,~ s=1 which gives  k=2 ~,~ l = 3 the only solution other than  (k,l) = (1,1) .
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