From: Kevin Brown [SMTP:ksbrown@seanet.com]
To: 'TORSTEN.SILLKE@LHSYSTEMS.COM'
Subject: RE: Heronian Triangles with Consecutive Integral Sides
Sent: 6/7/98 10:17 PM
V(n+1) = 4V(n) - V(n-1), V(0)=2, V(1)=4
>Conjecture1: If p is a prime divisor of V(n) - 1 then
>
> 1, 13 (mod 24) if n is even
> p =
> 1, 17 (mod 24) or p = 3 if n is odd
If n is even, we know there are integers a,b such that
V(n)-1 = a^2 - 3 = 1 + 3b^2
Notice that these are special cases of the quadratic forms
x^2 + Ny^2
where gcd(x,Ny)=1. Euler (and probably Fermat) proved that,
for the cases N=1, +-2, 3, every prime divisor of this form
is also expressible in this form, and the prime p can divide
a number of this form if and only if (-N) is a square modulo
p. Thus, any prime divisor of a number of the form a^2 - 3
(with a prime to 3) must be congruent to 1, 11, 13, or 23
modulo 24, because those are the only primes modulo which 3
is a square. Likewise, any prime divisor of a number of the
form 1 + 3b^2 must be congruent to 1, 7, 13, or 19 modulo 24,
because those are the prime modulo which -3 is a square. Now,
since V(n)-1 must be of BOTH those forms if n is even, it's
clear that any prime divisor must be in the intersection of
those two two sets of residues, so it can only be 1 or 13
(mod 24).
For the case of odd n we know there are integers a,b such that
V(n)-1 = 3(2a^2 - 1) = 2b^2 + 1
so in this case, after factoring out the prime 3, we need only
consider the prime divisors of numbers that have BOTH the forms
x^2 - 2y^2 and x^2 + 2y^2
with x coprime to 2y. It follows that any prime divisor of
2a^2 - 1 must be a prime p modulo which 2 is a square, so p
must be congruent to 1, 7, 17, or 23 (mod 24). Also, any
prime divisor of 2b^2 + 1 (other than 3) must be a prime p
modulo which -2 is a square, so p must be congruent to 1, 3,
11, 17, or 19 (mod 24). Again, since V(n)-1 must be of BOTH
those forms if n is odd, it's clear that any prime divisor
must be in the intersection of those two two sets of residues,
so it can only be 1 or 17 (mod 24) (once we have factored out
the power of 3).
Regards,
Kevin Brown
------
By the way, there is some discussion of the period lengths
of general linear recurrences modulo p in sections 2.7 and
2.8 of the article entitled "Symmetric Pseudoprimes" on my
web page (www.seanet.com/~ksbrown/). Also, there is an
article discussing congruence conditions on linear recurrences
of order 2, 3, and 4. You can also find an extensive
discussion of Lucas sequences in Ribenboim's "Book of
Prime Number Records", although I don't know if it contains
answers to any of your conjectures.
(By the way, it's interesting that this is the same Pell
equation that was apparently used to give Archimedes famous
bounds on sqrt(3).)