- View Source--- In primenumbers@yahoogroups.com,

"paulunderwooduk" <paulunderwood@...> wrote:

> Francois Arnault's paper, which I must learn

Here, as patiently as I am able, is how to fool any "tst(n,a)"

with, at is heart, in general terms, a Lucas test of type

L^((n+1)/2) = +/- 1 mod (n, L^2 - y*L +1)

with y some rational function of the parameter 'a' in terms

of which you have defined your effete additional

Fermat/Euler/M-R tests and gcd wriggles.

For some obscure reason, your last choice was

y = 2*(5*a^2-1)/(3*a^2+1). Now you have made a new choice.

But who cares? The fooling method would be the same.

OK, Paul? Have you got that? All we care about,

for Lucas, are powers of L mod (n, L^2 - y*L + 1).

Then here is the trick.

Chose some small odd number k > 1 (like k = 19 in my last hints)

and semiprimes of the form n=p*q with primes p = -1 mod k

and q = 1 + 2*k*m*(p-1), where m is small. (I usually choose m=1.)

Then find the irreducible polynomial P(x), with degree

eulerphi(k)/2, satisfied by 2*cos(2*Pi/k), and solve

for P(+/-y) = 0 mod n. For k = 19, P(x) is

? print(algdep(2*cos(2*Pi/19),9))

x^9 + x^8 - 8*x^7 - 7*x^6 + 21*x^5 + 15*x^4 - 20*x^3 - 10*x^2 + 5*x + 1

Exercise: Show that with x = +/- 2*(5*a^2-1)/(3*a^2+1),

the numerators give

{smart(a,n)=local(v=[

581130733*a^18-1549681961*a^16+1598109492*a^14-815197364*a^12

+219605318*a^10-31108446*a^8+2189796*a^6-67524*a^4+693*a^2-1,

290565367*a^18-920123659*a^16+1150154588*a^14-728431196*a^12

+250797682*a^10-47124218*a^8+4637292*a^6-217740*a^4+4047*a^2-19]);

...

as in the Gremlins latest totally successful attack.

Now all we have to do, for a given prime pair (p,q), is to

find the roots of these bizarre polys, mod n, using

polrootsmod and the CRT.

Up to effete signs, we have solved the Lucas problem.

Up to effete signs, we have solved all the additional

Fermat/Euler bolt-ons, mod p. All that is left are the

Fermats mod q. As is clear from Arnault, you will

not need to spend long looping on p, k, m, to solve

Fermats mod q = 1 + 2*k*m*(p-1), for at least one

of the many chinese roots.

That's it. All that can happen is that the deviser of the

hopeless test may wriggle with gcds, and arbitrarily bolt on

additional Fermats/Euler/M-R's, thereby filling PrimeNumbers

with lots of noise that cannot change the fact that she/he is

bound to lose, because, in your own words:

> one parameter Lucas plus N Fermat/Euler/M-R PRP test

End of story? End of long repetitious posts?

> can be counterexampled

David - View SourceTao, Harcos, Englesma, et al seem to have stalled trying to improve Zhang's upper

bound of 70,000,000. They claim to have confirmed they got it down to 5414 but look

like they aren't going to be able to go much further (perhaps can push it a bit below 5000

if combine all their juice?).

http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes