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• ## Re: Lucas super-pseudoprimes for Q <> 1

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• Back in May, in http://tech.groups.yahoo.com/group/primenumbers/message/21508 ... (David gave a nice example of a large n where condition (c) meant that Q was
Message 1 of 46 , Oct 30, 2010
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Back in May, in
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
• ... http://physics.open.ac.uk/~dbroadhu/cert/dbmo116.out gives my 116, in the format [n, factors, number of solutions] With n
Message 46 of 46 , Nov 9, 2010
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"mikeoakes2" <mikeoakes2@...> wrote:

> > My revised count up to 2*10^10 is 116.
> My (original) count up to 2*10^10 was 105.
> So it must have missed 11, i.e. a bigger proportion.