Re: Primes and Expanded Pythagorean Triples of the form (a+b)(a-b)^p-1=c^p

On Jul 7, 10:05 pm, rer <reri...@xxxxxxxxx> wrote:
On Jul 5, 11:31 pm, rer <reri...@xxxxxxxxx> wrote:



My apologies for the crossover between this and another post.

In everyPythagoreantriple one of the three triples is always
divisible by 5 (p=2, 2p+1=5). And, of course, 5 is prime.

Similarly, the proposition for the "expanded"Pythagoreantriples is,
if one of the three triples is always divisible by 2p+1, then 2p+1 is
prime. However, if one of them is not always divisible by 2p+1, then 2p
+1 is not prime.

Below are some examples.

Once again, since the numbers get large in a hurry, I am not able to
test very far. So, anyone willing to see if this also fails quickly?
(if this is not very readable, it is also available here --http://mysite.verizon.net/~reriker/exppythag-primes.html)

Someone was nice enough to run the following Mathematica code that I
turned into this file -- Warning! Warning! Warning! It's a relatively
large file at 4+ Mb:

http://mysite.verizon.net/reriker/10003n.htm

Mathematica code:
Table[
Table[a = Floor[c^p/2] + 1; b = a - 1;
{p, 2p+1, c, Mod[a, 2p+1], Mod[b, 2p+1], Mod[c, 2p+1]},
{c, 3, 69, 2}
],
{p, 2, 1000}
]// MatrixForm

I have not had a chance to look at every p/2p+1. However, unlike when
looking at just 3 for p < 1000, it does *not* show false positives at
60, 351, 770, 864, or 945.

Granted, it's still a relatively small sample, but it certainly looks
very promising.

rer


It is not as elegant as the solution for Pythagorean triples, but I
have found an equation that I believe gives all the integer solutions
for (a+b)(a-b)^p-1=c^p:

Let v, u be positive integers such that v>u

Then c=v+u

If Mod(c,v-u) is 0, then a & b have integer solutions such that

a=(v+u)^p+(v-u)^p/2(v-u)^(p-1)

b=a-(v-u)

I have updated the page with examples -- http://mysite.verizon.net/reriker/exppythag-primes.html
..

It appears to still be consistent with the postulate about primes.

.