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Re: n^n+ (n+1)^(n+1)

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  • mikeoakes2
    ... I believe the Lemma, but not my proof, because:- Take e.g. n=8; then the repeated factor of n^n-(n-1)^(n-1) is 19; but 19 does not divide (12*n^2+6*n+1). I
    Message 1 of 48 , Sep 6, 2009
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      --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
      >
      > Lemma: For positive integer n, the integer
      > f(n) = (6*n+2)^(6*n+2) - (6*n+1)^(6*n+1) is never square-free.
      >
      > Proof: Let w = (-1+sqrt(-3))/2. Then 1+w = -w^2 and
      > 3*(12*n^2+6*n+1) = (6*n+2+w)*(6*n+1-w).
      > Let P be a rational prime divisor of 12*n^2+6*n+1 and let
      > p be a corresponding prime divisor of 6*n+2+w in Q(sqrt(-3)), with
      > norm(p) = P. Then 6*n+2 = -w+m*p, for some m in Q(sqrt(-3)), and
      > f(n) = (-w+m*p)^(6*n+2) - (w^2+m*p)^(6*n+1).
      > Now we expand in p, using the binomial theorem, to obtain
      > (-w+m*p)^(6*n+2) = (-w)^(6*n+2)*(1+m*p) mod p^2 and
      > (w^2+m*p)*(6*n+1) = (w^2)^(6*n+1)*(1+m*p) mod p^2.
      > Since (-w)^(6*n+2) = (-w)^2 and (w^2)^(6*n+1) = w^2, we obtain
      > f(n) = 0 mod p^2. Hence f(n) is divisible by norm(p^2) = P^2
      > and is hence not square-free.

      I believe the Lemma, but not my proof, because:-
      Take e.g. n=8;
      then the repeated factor of n^n-(n-1)^(n-1) is 19;
      but 19 does not divide (12*n^2+6*n+1).
      I can't see the error in my algebra [am short of sleep].
      Please help.

      Mike
    • djbroadhurst
      ... This is now a conjecture-free OEIS sequence: http://www.research.att.com/~njas/sequences/A165284 David
      Message 48 of 48 , Sep 14, 2009
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@> wrote:

        > primes p in
        > http://www.research.att.com/~njas/sequences/A068209
        > for which (x+1)^p-x^p-1 and x^x+(x+1)^(x+1)
        > are never simultaneously divisible by p^2.

        This is now a conjecture-free OEIS sequence:
        http://www.research.att.com/~njas/sequences/A165284

        David
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