## Re: "wriggly" probable primes

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• ... 4065702994722252685573484796054334194691713593576645739409115721859519 = 5x^2 + 5xy + y^2 Comment: Pari-GP s qfbsolve enables a solution in two minutes.
Message 1 of 50 , Sep 14, 2010
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>> Exercise: Find two pairs of positive integers (x,y) such that
4065702994722252685573484796054334194691713593576645739409115721859519
= 5x^2 + 5xy + y^2
Comment: Pari-GP's "qfbsolve" enables a solution in two minutes.
Devotees of "issquare", like Aldrich, may take considerably longer.<<

Here is the 2 minute solution:

R(v)=local(a=G(v[1]),b=G(v[2]));vecsort([X(a*b),X(a*conj(b))],1);
G(p)=qfbsolve(Qfb(5,5,1),p)*[2+u,1]~;
X(z)=vecsort([Z(z),Z(z*u^2),Z(z/u^2)],1)[1];
Z(z)=local(t=z*sign(imag(z)),a=real(t),b=imag(t));[b,max(a-2*b,-a-3*b)];
{A=4065702994722252685573484796054334194691713593576645739409115721859519;
for(k=1,2,print("x="S[k][1]);print("y="S[k][2]));print(T" seconds");}

x=8742252927800705984069016007531927
y=42652016277694077218687373764822402
x=24507765935843939778835994638318507
y=8131530641557000519732930155331538
119 seconds

David
• ... [4] is meaningless, as it stands. You should write a double mod: 4. (1+x)^p = 1-x mod(x^2-a,p) ... There is no reason whatsoever to believe that [1] to [4]
Message 50 of 50 , Sep 29, 2011
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"bhelmes_1" <bhelmes@...> wrote:

> 1. let a jacobi (a, p)=-1
> 2. let a^(p-1)/2 = -1 mod p
> 3. a^6 =/= 1 mod p
> 4. (1+sqrt (a))^p = 1-sqrt (a)

[4] is meaningless, as it stands.
You should write a double mod:

4. (1+x)^p = 1-x mod(x^2-a,p)

> 1. Is it possible that there are other exceptions

There is no reason whatsoever to believe that
[1] to [4] establish that p is prime. Morevoer,
some folk believe that, for every epsilon > 0,
the number of pseudoprimes less than x may
exceed x^(1-epsilon), for /sufficiently/ large x.

> 2....
> there is a cyclic order ...

> 3....
> there is a cyclic order ...

The group of units (Z/nZ)* is /not/ cyclic
if n has at least two distinct odd prime fators.

David
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