--- In

primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> Counterexample with 25 digits!

Indeed there seems to be little obstacle to manufacturing

"Fermat plus Lucas" pseudoprimes that are much larger,

given Bernhard's unwise choice of a positive Kronecker symbol.

For example this 62-digit counterexample was found in 7 seconds,

which is less time than it takes to factor p by brute force:

{g=9982012609162784317210263215191345125837484159508891963337295;

a=4577317655410878067033652962262995530005977311539811603281218;

p=10549248000000000000000800652325984000000000015191702458402471;

if(p%4 == 3 && !issquare(p) && !issquare(a) &&

kronecker(a,p) == 1 && Mod(a,p)^((p-1)/2) == 1 &&

Mod((1+x)*Mod(1,p),x^2-a)^((p-1)/2) == g*x &&

Mod(g,p)^2*a == 1 && !isprime(p),

print("Counterexample with "#Str(p)" digits!"));}

Counterexample with 62 digits!

David (per proxy the pseudoprime gremlins)