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

Expand Messages
• ... Using 4 days pari-GP at 3.6Ghz I have solidified these results. For b1, b2, q
Message 1 of 81 , Oct 3, 2011
>
>
> > > Puzzle 3a: Find primes b1 < p1 < q, b2 < p2 < q, b1 < b2, s.t.
> > > b1^p1 = b2^p2 = 1 mod q^2, (with no limits on q^2).
> >
> > Here are 5 such pairs of triples:
> >
> > [132529, 277691, 555383]
> > [211811, 277691, 555383]
> >
> > [134507, 883703, 1767407]
> > [720901, 883703, 1767407]
> >
> > [292933, 1051553, 2103107]
> > [385079, 1051553, 2103107]
> >
> > [434243, 613189, 2452757]
> > [470711, 613189, 2452757]
> >
> > [1161143, 3700283, 7400567]
> > [2768789, 3700283, 7400567]
>
> and a sixth, also with p1 = p2:
>
> [ 417961, 5699017, 22796069]
> [1997161, 5699017, 22796069]

Using 4 days pari-GP at 3.6Ghz I have solidified these results.

For b1, b2, q < 10^7.5, the complete list of such q is:-
555383
1767407
2103107
2452757
7400567

12836987
14668163
15404867
16238303
19572647
22796069 [found by DB]
25003799
26978663
27370727

Each (q-1) is of the form 2^m*p, and p1 = p2 in each case.

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
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.