- --- In primenumbers@yahoogroups.com,

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

> > Puzzle 3a: Find primes b1 < p1 < q, b2 < p2 < q, b1 < b2, s.t.

and a sixth, also with p1 = p2:

> > 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]

[ 417961, 5699017, 22796069]

[1997161, 5699017, 22796069]

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 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