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

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• ... Great result! I wish I could entirely get it, since things like norm(p) and Q(...) and using imaginary numbers in this way is beyond what I have
Message 1 of 48 , Sep 4, 2009
>
> 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 and (-w^2)^(6*n-1) = -w, 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.
>

Great result! I wish I could entirely get it, since things like norm(p) and Q(...) and using imaginary numbers in this way is beyond what I have experienced or remember through the fog.

So, in simpleton terms, are we safe to say that for integer n > 0

(6n-2)^(6n-2) + (6n-1)^(6n-1)

has (12n^2 - 6n + 1)^2 as a factor?

And by inspection this case appears to be the only occurrence of square factors in numbers of the form a^a + (a+1)^(a+1).

Mark
• ... This is now a conjecture-free OEIS sequence: http://www.research.att.com/~njas/sequences/A165284 David
Message 48 of 48 , Sep 14, 2009