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

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• ... and a sixth, also with p1 = p2: [ 417961, 5699017, 22796069] [1997161, 5699017, 22796069] David
Message 1 of 81 , Sep 23, 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]

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
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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