Back in May, in

http://tech.groups.yahoo.com/group/primenumbers/message/21508
I wrote:

>

> Notation:

> For integers P and Q, and nonnegative integer n, define V(P,Q,n) by the recurrence relation

> V(P,Q,n) = P*V(P,Q,n-1)-Q*V(P,Q,n-2)

> with initial conditions V(P,Q,0)=2, V(P,Q,1)=P.

> Then as is well known V(P,Q,n)=x1^n+x2^n,

> where x1, x2 are the roots of the quadratic equation x^2-P*x+Q=0.

>

> Conjecture:

> If n is an odd composite integer, and V(P,Q,n) = P mod n for all P, for fixed Q>1, then

> (a) n is not a multiple of 3;

> (b) n is a Carmichael number, i.e. for all prime factors p of n, (p-1) divides (n-1);

> (c) Q is a multiple of every prime factor p of n such that (p+1) does not divide (n^2-1).

> Conversely, if (a) and (b) hold, and if we define Q_0 = product of every prime factor p of n such that (p+1) does not divide (n^2-1), then there is at least one Q which is a multiple of Q_0 and is such that V(P,Q,n) = P mod n for all P.

>

(David gave a nice example of a large n where condition (c) meant that Q was a multiple of 1.)

I have just finished a 1 GHz-yr investigation of this Conjecture, and have to report that it is (very slightly:-) false.

For every odd composite integer n < 10^6, not a multiple of 3, I found all Q < 10^6 satisfying the conditions of the Conjecture.

In every case but one, n was indeed a Carmichael number, but for n = 507529 = 11*29*37*43, the conditions are satisifed, for Q a multiple of 1247 = 29*43, yet n is not a Carmichael number, as (11-1) does not divide (n-1).

Is this surprising at all, David? (I've rather lost track of the theory after all this time).

Mike