## Re: puzzle for a counterexample

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• 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
Message 1 of 66 , Nov 1, 2012
Dear David,

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.

p = 1 mod 4

> > the improved test
> > 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.

Nice Greetings from the primes
a lot of beautifull flowers in the mathematical universe
Bernhard
• ... 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
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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