- --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

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

> --- In primenumbers@yahoogroups.com,

> "djbroadhurst" <d.broadhurst@> wrote:

>

> > > 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]

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 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