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2^n+1 = m^2

  1. Apr 30, 2005 #1
    Hi,
    I'm looking to solutions of: [tex]2^n+Q=m^2[/tex] , where [tex]Q=1[/tex] .
    Obviously, n must be odd.
    I already know the trivial solution: [tex]2^3+1=3^2[/tex] and I've started using a naive PARI/gp program for finding (n,m) up to n=59 . No success yet.
    Do you know about other solutions or about some theory ?

    This is related to Pell numbers [tex](P,Q)=(2,-1)[/tex] and to a series of Prime numbers studied by Newman, Shanks and Williams, called NSW numbers, and generated by: [tex](P,Q)=(6,1)[/tex] .
    The idea is to have [tex]D=P^2-4Q=2^n[/tex] and [tex]Q=\pm 1[/tex] .
    Since Mersenne numbers are square-free, (2,-1) is the unic solution for Q=-1.
    About Q=1, I don't know ...

    Tony
     
  2. jcsd
  3. Apr 30, 2005 #2

    Zurtex

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    Well I don't know the solution, but this is how I would look at it. First of all we are looking at odd numbers, although before we do I would like to mention:

    [tex]2^0 + 1 = 1^2[/tex]

    Right so the from we have is:

    [tex]2^{2m + 1} + 1 = 2 \cdot 2^{2m} + 1 = 2 \cdot \left( 2^m \right)^2 + 1 = 2a^2 + 1 = p^2[/tex]

    So you are looking for the integer solutions of:

    [tex]2 a^2 + 1 = p^2 \quad \text{where:} \quad a = 2^m \quad m, \, p \, \in \mathbb{N}[/tex]

    Now, it's not too difficult to deduce from that that p must be odd. Not only that but p - 1 must be twice another square number. Hence:

    [tex]p^2 = \left(2q + 1\right)^2 = 4q^2 + 4q + 1[/tex]

    So:

    [tex]2q^2 + 2q = a^2[/tex]

    [tex]q^2 + q = 2^{2m - 1}[/tex]

    [tex]q(q + 1) = 2^{2m - 1}[/tex]

    Which is a contradiction for q > 1. Now look at q = 1, which is p=3 and your solution. :smile:
     
    Last edited: Apr 30, 2005
  4. Apr 30, 2005 #3
    Thanks !

    Thanks ! That proves that (n,p)=(3,3) is the unic solution of [tex]2^n+1=p^2[/tex].
    Thanks to your proof, I think I have something shorter:
    [tex]2^n+1=p^2 \leftrightarrow 2^n=(p-1)(p+1)[/tex] .
    In order to have both [tex]p-1=2^\alpha[/tex] and [tex]p+1=2^\beta[/tex], it is clear that there is only 1 solution: [tex]p=3[/tex] .

    Thanks !
    Tony
     
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