--- In

primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

>

> This was the puzzle:

>

> > Puzzle: Find a composite positive odd integer n that passes

> > the Lucas test V(a,1,n) = a mod n, for every integer a.

>

>[snip]

>

> Then by googling

> > pseudoprime 7056721

> I arrived at

> http://adm.lnpu.edu.ua/downloads/issues/2008/N2/adm-n2-4.pdf

> which proves that n is a solution if and only if it is an odd

> square-free composite integer such that for each prime p|n

> n = +/- 1 mod p-1 ... [1]

> n = +/- 1 mod p+1 ... [2]

> This paper claims that 7056721 is the unique solution with n < 10^10.

David,

I have at last started to digest that paper.

In particular, I believe that the relation between Chebyshev and Lucas polynomials is:

T_n(x) = (1/2)*V(2*x,1,n)

Yes?

Their definition of a "Chebyshev number" is a composite number n such that

T_n(a) = a mod n

for all integers a.

This translates into:

V(2*a,1,n) = 2*a mod n

for all integers a.

That's a weaker requirement than yours (above), which needs the modular equality to hold for _odd_ first argument of V() as well.

I wonder whether or not it is really a weaker condition.

Any views on this?

Mike