## Re: Square factors of b^p-1

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• ... New experimentally-determined data point supports the picture:- 10^7.5 1944 2.489 Mike
Message 1 of 81 , Oct 3, 2011
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--- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
>
> >
> > Puzzle 2: Find a triplet of primes (b,p,q) with b < p < q,
> > b^p = 1 mod q^2, and q^2 having precisely 21 decimal digits.
>
> I reckon there should be about 208952 solutions to this puzzle.
> Rationale:-
>
> max_q solutions ratio
> 10^1 0 -
> 10^1.5 1 -
> 10^2 1 1.0
> 10^3 1 1.0
> 10^3.5 1 1.0
> 10^4 3 3.0
> 10^4.5 6 2.0
> 10^5 22 3.667
> 10^5.5 58 2.636
> 10^6 125 2.155
> 10^6.5 322 2.576
> 10^7 781 2.435
> Taking the ratio to be about 2.4 then gives these estimates:-
> 10^7.5 1874 2.4
>...

New experimentally-determined data point supports the picture:-
10^7.5 1944 2.489

Mike
• Hi , all , Not even draft yet , let alone published : http://upforthecount.com/math/nnp.html still needs a lot of work . But , I have updated the top of the
Message 81 of 81 , Nov 29, 2011
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Hi , all ,

Not even draft yet , let alone published :
http://upforthecount.com/math/nnp.html
still needs a lot of work .
But , I have updated the top of the main table :
http://upforthecount.com/math/nnp2.txt
with complete factorizations now extending through
np ( 90 ) .

When is np ( n ) prime ?
So far , only for 1 , 2 and 3 .
When n mod 6 = 4 , np ( n ) has at least 3 algebraic
factors , 3 and n^2 + n + 1 ( twice ) .
But what about the residue ( primitive part ? ) ?
When is this prime ?
n t n^n + (n+1)^(n+1)
4 4 3 * 7 * 7 * 23
16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651
Any others ?

Cheers ,

Walter
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