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## Re: Proposition for a better p-1 and p+1 test

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• ... No prime p = 1 mod 4 passes this test . David
Message 1 of 14 , Sep 10, 2013
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<bhelmes@> wrote:

> I propose the following unproven prime test

No prime p = 1 mod 4 passes this "test".

David
• ... As remarked, no prime p = 1 mod 4 passes the test. Here I show that the composite number p = 2953*5167 = 15258151 = 7 mod 16 passes the test, using its
Message 2 of 14 , Sep 11, 2013
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<bhelmes@> wrote:

> I propose the following unproven prime test

As remarked, no prime p = 1 mod 4 passes the test.
Here I show that the composite number
p = 2953*5167 = 15258151 = 7 mod 16
passes the test, using its minimal strong base, a = 3:

{p=15258151;x=13785839;y=8920588;xr=1472312;yr=6337563;
if(Mod(p,16)==7&&kronecker(2,p)==1&&kronecker(3,p)==-1&&
Mod(3,p)^((p-1)/2)==-1&&x^2-3*y^2==Mod(1,p)&&
Mod(Mod(1,p)*(x+y*A),A^2-3)^p==x-y*A&&
Mod(Mod(1,p)*(x+y*A),A^2-3)^((p+1)/8)==xr+yr*A&&
gcd(p,yr)==1&&!isprime(p),print(" fooled with minimal base"));}

fooled with minimal base

The gremlins will entertain no further wriggle from Bernhard.

David (their keeper)
• ... As remarked, no prime p = 1 mod 4 passes the test. Here I show that the composite number p = 2953*5167 = 15258151 = 7 mod 16 passes the test, using its
Message 3 of 14 , Sep 13, 2013
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Dear David,

thank you for your efforts to find a counterexample.

I noticed that you did not used a solution of pell for a=3 with x^2-ay^2=1

but the solution for x^2-ay^2=1 mod p

I am not sure wether this makes a difference for the test.

I wish you a pleasant and peaceful weekend

Bernhard

--- In primenumbers@yahoogroups.com, <david.broadhurst@...> wrote:

<bhelmes@> wrote:

> I propose the following unproven prime test

As remarked, no prime p = 1 mod 4 passes the test.
Here I show that the composite number
p = 2953*5167 = 15258151 = 7 mod 16
passes the test, using its minimal strong base, a = 3:

{p=15258151;x=13785839;y=8920588;xr=1472312;yr=6337563;
if(Mod(p,16)==7&&kronecker(2,p)==1&&kronecker(3,p)==-1&&
Mod(3,p)^((p-1)/2)==-1&&x^2-3*y^2==Mod(1,p)&&
Mod(Mod(1,p)*(x+y*A),A^2-3)^p==x-y*A&&
Mod(Mod(1,p)*(x+y*A),A^2-3)^((p+1)/8)==xr+yr*A&&
gcd(p,yr)==1&&!isprime(p),print(" fooled with minimal base"));}

fooled with minimal base

The gremlins will entertain no further wriggle from Bernhard.

David (their keeper)
• ... There is an infinity of solutions to x^2-3*y^2 = 1, with integers x and y. Here are a few: q=quadunit(12);r=1;for(k=1,20,r*=q;print([k,real(r),imag(r)]));
Message 4 of 14 , Sep 13, 2013
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Bernhard Helmes wrote:

> you did not used a solution of pell for a=3 with x^2-ay^2=1

There is an infinity of solutions to x^2-3*y^2 = 1,
with integers x and y.

Here are a few:

q=quadunit(12);r=1;for(k=1,20,r*=q;print([k,real(r),imag(r)]));

[1, 2, 1]
[2, 7, 4]
[3, 26, 15]
[4, 97, 56]
[5, 362, 209]
[6, 1351, 780]
[7, 5042, 2911]
[8, 18817, 10864]
[9, 70226, 40545]
[10, 262087, 151316]
[11, 978122, 564719]
[12, 3650401, 2107560]
[13, 13623482, 7865521]
[14, 50843527, 29354524]
[15, 189750626, 109552575]
[16, 708158977, 408855776]
[17, 2642885282, 1525870529]
[18, 9863382151, 5694626340]
[19, 36810643322, 21252634831]
[20, 137379191137, 79315912984]

How do you intend to prove that one pair of Pell's /infinite/
series does not correspond to my choice, modulo p?

That might take you some time :-?

David
• ... But it will not take for ever. It would be sufficient for Bernhard to study less than 960,000 integers from http://oeis.org/A001075 ... Of these, the
Message 5 of 14 , Sep 13, 2013
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--- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:

> How do you intend to prove that one pair of Pell's /infinite/
> series does not correspond to my choice, modulo p?
>
> That might take you some time :-?

But it will not take for ever. It would be sufficient for
Bernhard to study less than 960,000 integers from
http://oeis.org/A001075
> a(n) solves for x in x^2 - 3*y^2 = 1

Of these, the largest has less than 550,000 digits:

{stop=ceil(15258151/16);digs=#Str(real(quadunit(12)^stop));
print(stop"th Pell number has "digs" decimal digits");}

953635th Pell number has 545429 decimal digits

Happy hunting :-)

David
• Dear David, As remarked, no prime p = 1 mod 4 passes the test. I would like to add that only the point 5. causes the trouble. and that all primes passes the
Message 6 of 14 , Sep 14, 2013
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Dear David,

>As remarked, no prime p = 1 mod 4 passes the test.

I would like to add that only the point 5. causes the trouble.
and that all primes passes the points 1. - 4.

>Here I show that the composite number
>p = 2953*5167 = 15258151 = 7 mod 16
>passes the test, using its minimal strong base, a = 3:

I feel that it was not easy for you and a lot of work for you to find one counterexample.
Besides i noticed that your selected x and y is kind of  (x+yA)^4=-1 mod p
which is very special and shows that the proposed test is not sufficient.

>The gremlins will entertain no further wriggle from Bernhard.

David, i have a huge respect for your mathematical skills,
nevertheless the description to entertain me is not what i expect from you.
This primenumber group is the only place for me to discuss some mathematical
ideas and to get some mathematical feedback.

As far as i can see a proven sufficient primetest with 4 Selfridge would be a mathematical progress
and the development of choosing the right parameters in order to build a proof is not a kind of "wriggle" but
mathematical work.

I appreciate deeply a fair mathematical discussion.

David, prime numbers are something for the eternity and there is no reason to get angry about some humble
mathematicians like me who try to learn some more details of number theory.

I wish you a pleasant and friendly weekend.
Bernhard

David (their keeper)
• ... So remove point (5) to allow primes p = 1 mod 4. Then it s /very/ easy to find a pseudoprime: {p=3281;x=1062;y=1600; if(kronecker(3,p)==-1&&
Message 7 of 14 , Sep 14, 2013
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Bernhard Helmes wrote:

> > As remarked, no prime p = 1 mod 4 passes the test.
> I would like to add that only the point 5. causes the trouble.

So remove point (5) to allow primes p = 1 mod 4.
Then it's /very/ easy to find a pseudoprime:

{p=3281;x=1062;y=1600;
if(kronecker(3,p)==-1&&
Mod(3,p)^((p-1)/2)==-1&&
x^2-3*y^2==Mod(1,p)&&
Mod(Mod(1,p)*(x+y*A),A^2-3)^p==x-y*A&&
!isprime(p),print(" fooled"));}

fooled

David
• ... The gremlins now have more than 23,000 pseudoprimes that fool Bernhard s test. But before revealing them, they await a response to the very simple case,
Message 8 of 14 , Sep 20, 2013
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--- In primenumbers@yahoogroups.com,
"djbroadhurst" <david.broadhurst@...> wrote:

> {p=3281;x=1062;y=1600;
> if(kronecker(3,p)==-1&&
> Mod(3,p)^((p-1)/2)==-1&&
> x^2-3*y^2==Mod(1,p)&&
> Mod(Mod(1,p)*(x+y*A),A^2-3)^p==x-y*A&&
> !isprime(p),print(" fooled"));}
>
> fooled

The gremlins now have more than 23,000 pseudoprimes
that fool Bernhard's test. But before revealing them,
they await a response to the very simple case, above.

David
• Dear David, the amount of counterexamples sounds impressively. I suppose that there are a lot of p=1 mod 4. I think that i will publish soon a test which
Message 9 of 14 , Sep 21, 2013
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Dear David,

the amount of counterexamples sounds impressively.
I suppose that there are a lot of p=1 mod 4.
I think that i will publish soon a test which separates between p=1 mod 4 and p=3 mod 4,
but i would like to check first some list by myself in order to save some time for you and to get
some more informations.

How does it feel to be a "keeper" of such a lot of interesting gremlins :-)

A new and better proposition:

A. Let p an element of N with p=3 mod 4,
m the greatest power of 2 with 2^m*r=p+1
and 2^m < r

1. choose a prime a with jacobi (a, p)=-1
2. Verify a^(p-1)/2=-1 mod p
3. Depending on the a choose the pell solution of x, y with
x^2-ay^2=1 mod p
http://mathworld.wolfram.com/PellEquation.html
4. Verify that (x+yA)^(2^m)<>1 mod p  with A=sqrt (a)
because gcd (p-1, p+1) = 2
5. Calculate and verify (x+yA)^p=x-yA mod p
6. Calculate (x+yA)^r=x(r)+y(r)A mod p and
verify that  gcd (y(r), p)=1

(The point 6. derives of a factorisation test, is cheap and powerful
because gcd (y(n), p) > 1 results gcd (y(m), p) > 1 with m=n*t)

if 1.-6. is ok. then p is a prime.

David, you will realize that i abstain form the choose of the smallest a
and that i add the point 4. for good reasons

I wish you a peaceful and pleasant weekend
Bernhard
• ... The gremlins refuse to consider a test that is not true for primes. Consider, for example, the prime p = 1037*2^8-1 = 265471 and set a = 3, x = 2, y = 1.
Message 10 of 14 , Sep 21, 2013
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Bernhard Helmes wrote:

> A new and better proposition

The gremlins refuse to consider a
test that is not true for primes.

Consider, for example, the prime
p = 1037*2^8-1 = 265471
and set a = 3, x = 2, y = 1.

Then this prime fails your latest wriggle:

> (x+yA)^(2^m)<>1 mod p with A=sqrt(a)

print(Mod(Mod(1,265471)*(2+A),A^2-3)^(2^8));
Mod(Mod(1, 265471), A^2 - 3)

Please do not post probable primality "tests"
that are failed by primes.

David (trying to control the grumpy gremlins:-)
• A peaceful day for all group members, dear David, I changed only the point 3. Is this changement a sharper condition or is this mathematically seen the same ?
Message 11 of 14 , Oct 12, 2013
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A peaceful day for all group members,

dear David,

I changed only the point 3.

Is this changement a sharper condition or is this mathematically seen the same ?

I. For p=3 mod 4

1. choose the smallest prime a with jacobi (a, p)=-1
2. Verify a^(p-1)/2=-1 mod p
[This is the p-1 test which proves that a is a non quadric residuum]
3. Depending on the a choose the pell solution of x, y with
x^2-ay^2=-1 mod p (the minus is the change :-)
4. Calculate and verify (x+yA)^p=x-yA mod p with A=sqrt (a)
5. m the greatest power of 2 with 2^m*r=p+1 and 2^m < r

Calculate (x+yA)^r=x(r)+y(r)A mod p and
verify that  gcd (y(r), p)=1
[This is a strong p+1 test with a solution of the according pell equation]

if 1. - 5. is true then p should be a prime

I think that this "wriggle" will include that there exist a cycle group with order t > 2 and t | r

There is no need to hurry up for the gremlins,
the prime numbers will exist a little longer. ;-)

Friendly greetings from the primes
Bernhard

• ... For a=3 mod 4 there is no such thing. Please do not post tests that refer to non-existent entities. David
Message 12 of 14 , Oct 12, 2013
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Bernhard Helmes wrote:

> the pell solution of x, y with x^2-ay^2=-1

For a=3 mod 4 there is no such thing.

Please do not post "tests" that refer
to non-existent entities.

David

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