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## Re: mod quartic composite tests

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• ... Exercise 5: Show that the test loses 3 selfridges for 2*x^2 = 5 mod n, degenerating to a 1-selfridge Euler test, with base -15/16, plus a 2-selfridge Lucas
Message 1 of 26 , Jan 18, 2013
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> Exercise 4: Show that the test loses 2 selfridges for 2*x^2 = 5 mod n.

Exercise 5: Show that the test loses 3 selfridges for 2*x^2 = 5 mod n,
degenerating to a 1-selfridge Euler test, with base -15/16, plus a
2-selfridge Lucas test with P = 2/5 and Q = 1, and thus costs the same as BPSW.

Comment: As in the case of BPSW, the gremlins cannot defraud this case.

David
• I failed to solve any of David s exercises... but I have done some shallow verification, n
Message 2 of 26 , Jan 19, 2013
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I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:

{tst(n,x)=kronecker(x^2-4,n)==-1&&
gcd(x+1,n)==1&&
Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}

It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,

Paul -- subject to gcd wriggles
• ... Further, it seems that if gcd(x+1,n) is needed then it is equal to 1 (mod 6) Paul
Message 3 of 26 , Jan 19, 2013
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
>
>
> I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:
>
> {tst(n,x)=kronecker(x^2-4,n)==-1&&
> gcd(x+1,n)==1&&
> Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
>
> It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,
>

Further, it seems that if "gcd(x+1,n)" is needed then it is equal to 1 (mod 6)

Paul
• ... These ancillary statements are mostly false, except that maybe when gcd(x+1,n) needs to be checked then gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and
Message 4 of 26 , Jan 19, 2013
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
>
>
>
> --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
> >
> >
> > I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:
> >
> > {tst(n,x)=kronecker(x^2-4,n)==-1&&
> > gcd(x+1,n)==1&&
> > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
> >
> > It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,
> >
>
> Further, it seems that if "gcd(x+1,n)" is needed then it is equal to 1 (mod 6)
>

These ancillary statements are mostly false, except that maybe when "gcd(x+1,n)" needs to be checked then gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) are either 1 or prime,

Paul
• ... Please accept my apology for my previous statements about this composite test. I am actually running tests for: (mod n, L^2-x*L+1) and (mod n, L^2+x*L+1);
Message 5 of 26 , Jan 19, 2013
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

> > >
> > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
> > > gcd(x+1,n)==1&&
> > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
> > >

Please accept my apology for my previous statements about this composite test. I am actually running tests for:
(mod n, L^2-x*L+1) and (mod n, L^2+x*L+1); not the product of these. If the main test is passed then gcds with n of x-1, x, x+1, x^2-2 and x^2-3 are logged.

Now for some speculation about the results so far:

1) taking the mod with "the product" implies gcd(x,n)==1.

2) for "the product", gcds of n with x-1, x^2-2 and x^2-3 are either 1 or a prime congruent to 5 (mod 6) where "gcd(x+1,n)" is logged.

3) logged gcd(x+1,n) is not 1

4) the logged n are all congruent to 5 (mod 6).

Paul
• ... Here is another test, on the same theme, for which I cannot also easily find a fraud: {tst(n,x)=kronecker(x^2-4,n)==-1&& gcd(x^2-1,n)==1&&
Message 6 of 26 , Jan 21, 2013
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

>
> > > >
> > > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
> > > > gcd(x+1,n)==1&&
> > > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
> > > >
>

Here is another test, on the same theme, for which I cannot also easily find a fraud:

{tst(n,x)=kronecker(x^2-4,n)==-1&&
gcd(x^2-1,n)==1&&
Mod(Mod(1,n)*(L+x^2-1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-1;}

Paul
• ... n=2953711;x=285843 is a near-counterexample that comes from me testing over the two quadratics that form the quartic. gcd(x,n)==95281 and
Message 7 of 26 , Jan 27, 2013
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

> Here is another test, on the same theme, for which I cannot also easily find a fraud:
>
> {tst(n,x)=kronecker(x^2-4,n)==-1&&
> gcd(x^2-1,n)==1&&
> Mod(Mod(1,n)*(L+x^2-1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-1;}
>

n=2953711;x=285843 is a near-counterexample that comes from me testing over the two quadratics that form the quartic. gcd(x,n)==95281 and gcd(x^2-2,n)==31,

Paul
• ... n=1934765;x=1219266 has gcd(x-1,n)==265. So 2) might be 1 or a product of primes each congruent 5 (mod 6) , but as greater n get tested I guess this rule
Message 8 of 26 , Jan 30, 2013
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> --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
>
>
> > > >
> > > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
> > > > gcd(x+1,n)==1&&
> > > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
> > > >
>
> Please accept my apology for my previous statements about this composite test. I am actually running tests for:
> (mod n, L^2-x*L+1) and (mod n, L^2+x*L+1); not the product of these. If the main test is passed then gcds with n of x-1, x, x+1, x^2-2 and x^2-3 are logged.
>
> Now for some speculation about the results so far:
>
> 1) taking the mod with "the product" implies gcd(x,n)==1.
>
> 2) for "the product", gcds of n with x-1, x^2-2 and x^2-3 are either 1 or a prime congruent to 5 (mod 6) where "gcd(x+1,n)" is logged.
>

n=1934765;x=1219266 has gcd(x-1,n)==265. So 2) might be "1 or a product of primes each congruent 5 (mod 6)", but as greater n get tested I guess this rule will break too...

> 3) logged gcd(x+1,n) is not 1
>
> 4) the logged n are all congruent to 5 (mod 6).
>

I have verified all n<1.95*10^6

Paul
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