- Dear David,

thank you for your efforts,

i have the slight suspision that there may be counterexamples

for n=3 mod 4 but not for n=1 mod 4

even if i could not prove it.

I know that you are very fast in replying.

If you find one counterexample of the form n=1 mod 4

i would be glad to know it.

Sorry that i bother you with such a lot of questions,

but i hope that there will be some success.

Take your time.

p = 1 mod 4

> > the improved test

Nice Greetings from the primes

> > 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.

a lot of beautifull flowers in the mathematical universe

Bernhard - --- In primenumbers@yahoogroups.com,

paulunderwooduk" <paulunderwood@...> wrote:

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

Here are some scores out of 5:

> should be -1,1,0,-2, or 2

{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