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

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• ... 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].

Mike
• ... 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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