## Re: puzzle for a counterexample

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• ... In 3.5 GHz-hours, the gremlins found 33,445 counterexamples. Since Kermit likes big files of semiprimes, the results are in
Message 1 of 66 , Nov 2, 2012
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"bhelmes_1" <bhelmes@> wrote:

> 1. gcd (a, p)=1
> 2. Let jacobi (a, p) = -1
> 3. let jacobi (a^2-a, p)=-1
> 4. a^[(p-1)/2]=-1 mod p
> 5. if (a+sqrt (a))^p = a-sqrt(a) mod p
> 6. s+t*sqrt(a):=(a+sqrt (a))^[(p+1)/2] implies that
> gcd (s, p)=1 or 0 or p, and gcd (t, p)=1 or 0 or p.

In 3.5 GHz-hours, the gremlins found 33,445 counterexamples.

Since Kermit likes big files of semiprimes, the results are in

They all fool Bernhard's 4-selfridge test, above, with my
strengthening: s = 0 mod p, gcd(t,p) = 1, as shown here:

{tst(p,a)=local(u,v,s,t);
if(gcd(a,p)==1&&kronecker(a,p)==-1&&
kronecker(a^2-a,p)==-1&&Mod(a,p)^((p-1)/2)==-1,
u=Mod(Mod(1,p)*(a+x),x^2-a);
if(u^p==a-x,v=lift(lift(u^((p+1)/2)));
s=polcoeff(v,0);t=polcoeff(v,1);
if(s||gcd(t,p)-1,t=0)));t;}

for(k=1,#v,n=v[k][1];a=v[k][2];
if(tst(n,a)&&!isprime(n),c++;if(n%4==1,d++)));
print1(c" counterexamples in "file" : ");
print(round(100*d/c)"% with n = 1 mod 4");}

tstfile("underbh4.txt");

33445 counterexamples in underbh4.txt : 46% with n = 1 mod 4

David (per proxy the industrious gremlins)
• ... Here are some scores out of 5: {A(k,x)=sum(j=0,k/2,(-1)^j*binomial(k-j,j)*x^(k-2*j));} {B(k,x)=sum(j=0,(k-1)/2,(-1)^j*binomial(k-j-1,j)*x^(k-2*j-1));}
Message 66 of 66 , Nov 22, 2012
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paulunderwooduk" <paulunderwood@...> wrote:

> At least one of the evaluations of x at -1,1,0,-2 or 2
> should be -1,1,0,-2, or 2

Here are some scores out of 5:

{A(k,x)=sum(j=0,k/2,(-1)^j*binomial(k-j,j)*x^(k-2*j));}
{B(k,x)=sum(j=0,(k-1)/2,(-1)^j*binomial(k-j-1,j)*x^(k-2*j-1));}
{L=[-1,1,0,-2,2];S=Set(L);for(k=2,40,f=factor(A(k,x)-B(k,x))[,1];
g=f[#f];c=0;for(j=1,#L,if(setsearch(S,subst(g,x,L[j])),c++));
print1(c","));}

4,4,3,4,4,4,4,4,4,4,4,3,4,4,4,5,4,4,4,4,5,4,4,4,4,5,4,4,4,5,5,4,4,4,4,4,5,4,3,

David
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