14° Primality test and factorization of Lepore ( conjecture )

[Alberico Lepore]

Originally Posted by CRGreathouse

If it depends on the number, it's not a constant. What kind of dependency does it have?

if I'm not mistaken if N = a ^ 2 + n * a without repetitions goes from 1 to a / (2 * n)

[LaurV]

Originally Posted by Alberico Lepore

if I'm not mistaken if N = a ^ 2 + n * a without repetitions goes from 1 to a / (2 * n)

Do you mean you have an algorithm which checks if N=a*(a+n) is prime?

[Alberico Lepore]

Originally Posted by LaurV

Do you mean you have an algorithm which checks if N=a*(a+n) is prime?

I had never thought about it.
Give me a few minutes to think about it

[Alberico Lepore]

Originally Posted by LaurV

Do you mean you have an algorithm which checks if N=a*(a+n) is prime?

thank you,
you pointed out to me that I was wrong
it is 2 ^ X, X with repetitions
but I do not give up

[Alberico Lepore]

The algorithm that I show you is in X with repetitions, if you want to apply the variant of log_2, you will have X * log_2 () with X without repetitions.

For Z we are interested if you admit solutions or not
For others we must test whether the parent's product difference is the same but opposite sign from that of the child
Example

CHECK Z

[1135793-2*a^2]=-Z , Z=-a*(a-n) , a^2+n*a=1135793 YES solution bad way , Z is positive

[1135793-2*a^2]=Z , Z=-a*(a-n) , a^2+n*a=1135793 NO solution good way


CHECK Q

[1135793-2*a^2]=Z , Z=-a*(a-n), Z-2*(a)^2=-[a*[(a)-(-(a-n)-(a))]]=A , a^2+n*a=1135793

[1135793-2*a^2]=Z , Z=-a*(a-n), Z-2*(a-n)^2=-[(-(a-n))*[-(a-n)-(a-(-(a-n)))]]=B , a^2+n*a=1135793

[-[(a)-(-(a-n)-(a))]-a]=-((a-n)-(a)) , a^2+n*a=1135793 There are no solutions are different then Q = A

-[a*[(a)-(-(a-n)-(a))]]=a*(-(3*a-n))=Q


CHECK R

Q-2*a^2=[a*(-(3*a-n))]-2*a^2=-[a*[a-(-(3*a-n)-a)]]=A , a^2+n*a=1135793

Q-2*(-(3*a-n))^2=-[(-(3*a-n))*[(-(3*a-n))-(a-(-(3*a-n)))]]=B , a^2+n*a=1135793

[-[a-(-(3*a-n)-a)]-a]=-[[(a)-(-(a-n)-(a))]-a] , a^2+n*a=1135793 There are no solutions are different then R = A

-[a*[a-(-(3*a-n)-a)]]=a*(-5*a+n)=R


CHECK T

R-2*a^2=-[a*[a-((-5*a+n)-a)]]=A , a^2+n*a=1135793

R-2*(-5*a+n)^2=-[(-5*a+n)*[(-5*a+n)-(a-(-5*a+n))]]=B , a^2+n*a=1135793

[(-5*a+n)-a]=-[[a-((-5*a+n)-a)]-a] , a^2+n*a=1135793 TRUE they are the same, better to say they are opposite then T = B

-[(-5*a+n)*[(-5*a+n)-(a-(-5*a+n))]]=(-5*a+n)*(-11*a+2*n)=T


etc.etc.




Does anyone confirm me the accuracy of the algorithm?