- Not much action on two of the conjectures,

and time grows short, so I will do the

obvious and increase the purse. Maybe even

DB himself will swoop out of the clouds to

pounce on this one! Of course, it just may

be possible that there really are no

counterexamples and the durn things are

impossible to prove. That would be unfortunate.

I offer a $100 prize to the first person who can submit

a verifiable counterexample or proof by New Year's day

for the following primality conjecture:

(x,A,B,c,k,f : integers)

Let A = 20x^2 + 10x + 1;

Let B = 10x^2 + 4x + 1;

Let c = trunc(A/sqrt(5)) - 1;

For any x > 0, apply the issquare test to each k

in the interval c <= k < B. If there is no value

of k that satisfies the conditions of the test,

then A is Prime.

(issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

I offer a $101 prize to the first person who

can submit a verifiable counterexample or proof

by New Year's day for the following conjecture:

(x,A,B,c,k,f : integers)

Let A = 20x^2 + 10x + 1;

Let B = 10x^2 + 4x + 1;

Let c = trunc(A/sqrt(5)) - 1;

For any x > 0, apply the issquare test to each k

in the interval c <= k < B.

If there exists a value of k that satisfies the

conditions of the test, then A is composite,

and there will be at least one value of k such

that 2*k - 1 will have a factor in A.

(issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

Aldrich Stevens - Happy new year to all

> I offer a $100 prize to the first person who can submit

I provide counterexamples to the contrapositive:

> a verifiable counterexample or proof by New Year's day

> for the following primality conjecture:

> (x,A,B,c,k,f : integers)

> Let A = 20x^2 + 10x + 1;

> Let B = 10x^2 + 4x + 1;

> Let c = trunc(A/sqrt(5)) - 1;

> For any x > 0, apply the issquare test to each k

> in the interval c <= k < B. If there is no value

> of k that satisfies the conditions of the test,

> then A is Prime.

> (issquare test: 0 < 5*(2*k -1)^2 - 4*A^2 = f^2)

If A is composite then there exists a k in the interval that satisfies the

test.

The maple code:

*> for x from 1 to 9 do> A:=20*x^2+10*x+1;

*

> if not isprime(A) then

> B:=10*x^2+4*x+1;

> c:=floor(A/sqrt(5))-1;

> printf("%A %A-%A ", x, c, B-1);

> for k from c to B-1 do

> y := 5*(2^k-1)^2-4*A^2;

> if y<=0 then next; end if;

> if issqr(y) then printf("(%A)", k); break; end if;

> end do;

> printf("\n");

> end if;

> end do:

The output:

4 160-176

5 245-270

6 348-384

9 764-846

Each row begins with a value x for which A is composite.

It is followed by the range of k. That is from c to B.

The code is written such that by finding a k value it is printed in ().

But there is no (k)

So we have at least 4 counterexamples trivially.

Or I have misunderstood the problem

Regards

Payam

[Non-text portions of this message have been removed] - Hello

Aldrich Stevens noted offline that my result is not correct:

> There must be an error in your program.

and Peter Kosinar again offline located the my mistake:

> All of your proposed counterexamples

> actually do have values of k that

> satisfy the conditions of the test.

> x = 4, k = 162

> x = 5, k = 247

> x = 6, k = 351, 358

> x = 9, k = 769

>> (issquare test: 0 < 5*(2*k-1)^2 - 4*A^2 = f^2)

Thank you Aldrich and Peter

>> y := 5*(2^k-1)^2 - 4*A^2;

> It seems your code uses 2^k, while the original poster used 2*k.

Changing the code as Peter noted

and adding a single dot after 5 in sqrt(5) to avoid maple's symbolic

manipulations

and extending the for x to higher values

the output reads as follows:

4 160-176 (162)

5 245-270 (247)

6 348-384 (351)

9 764-864 (769)

10 938-1040 (952)

...

100 89889-100400 (89893)

...

1000 8948743-10004000 (8959147)

...

Yes, not the conjecture nor its inverse is trivial!

Regards,

Payam

[Non-text portions of this message have been removed]