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

Expand Messages
• ... These have larger q: 1783447^48649379 = 1 mod 97298759^2 25659449^48649379 = 1 mod 97298759^2 2076259^48687659 = 1 mod 97375319^2 34543973^48687659 = 1 mod
Message 1 of 81 , Oct 3, 2011
• 0 Attachment

> ... the most lucrative case, p1=p2=(q-1)/2 ...
>
> 25638913^35364419 = 1 mod 70728839^2
> 26450899^35364419 = 1 mod 70728839^2

These have larger q:

1783447^48649379 = 1 mod 97298759^2
25659449^48649379 = 1 mod 97298759^2

2076259^48687659 = 1 mod 97375319^2
34543973^48687659 = 1 mod 97375319^2

David
• 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
• 0 Attachment
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
Your message has been successfully submitted and would be delivered to recipients shortly.