Re: n^n+ (n+1)^(n+1)
- --- In firstname.lastname@example.org, "djbroadhurst" <d.broadhurst@...> wrote:
>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.
> 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.
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).
- --- In email@example.com,
"djbroadhurst" <d.broadhurst@> wrote:
> primes p inThis is now a conjecture-free OEIS sequence:
> for which (x+1)^p-x^p-1 and x^x+(x+1)^(x+1)
> are never simultaneously divisible by p^2.