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• ## Re: [PrimeNumbers] Update on CONTEST! and CONTEST++ \$\$

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• Happy new year to all ... I provide counterexamples to the contrapositive: If A is composite then there exists a k in the interval that satisfies the test. The
Message 1 of 3 , Dec 28 11:36 AM
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Happy new year to all

> 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 provide counterexamples to the contrapositive:
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

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• Hello ... Thank you Aldrich and Peter Changing the code as Peter noted and adding a single dot after 5 in sqrt(5) to avoid maple s symbolic manipulations and
Message 1 of 3 , Dec 29 2:28 AM
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Hello

Aldrich Stevens noted offline that my result is not correct:

> There must be an error in your program.
> 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

and Peter Kosinar again offline located the my mistake:

>> (issquare test: 0 < 5*(2*k-1)^2 - 4*A^2 = f^2)
>> y := 5*(2^k-1)^2 - 4*A^2;
> It seems your code uses 2^k, while the original poster used 2*k.

Thank you Aldrich and Peter

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

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

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