## 25241Re: 4 Fermat and 1 Lucas [freely admitted by its author to be hopeless]

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• Jul 19, 2013
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"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
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