Matrix sieve theorem [closed]

Question

I have formulated the following conjecture:

Odd positive integer $ N=6n-1$ is a prime number iff neither of two diophantine equations

$6x^2+(6x−1)y=n$

$6x^2+(6x+1)y=n$

has solution. $x=1,2,3,..y=0,1,2,...n =1,2, 3..$

Odd positive integer $ N=6n+1$ is a prime number iff neither of two diophantine equations

$6x^2−2x+(6x−1)y=n$

$6x^2+2x+(6x+1)y=n$

has solution. $x=1,2,3,..y=0,1,2,...n =1,2, 3..$

Note: all primes, except 2 and 3, belong to two sequences: $S1(n)=6n-1$ or $S2(n)=6n+1$ Theorem allows to substitute the task: "Find all primes in the range $(N_1;N_2)$" by the task: "Find positive integers which do not appear in the range $(n_1;n_2)$ in two pairs of $2$-dimensional arrays

$P_1(i,j)=6i^2+(6i-1)(j-1)$

$P_2(i,j)=6i^2+(6i+1)(j-1)$

$i,j = 1,2,3,\ldots$

for primes in the sequence $N=6n-1$.

$P_3(i,j)=6i^2-2i+(6i-1)(j-1)$

$P_4(i,j)=6i^2+2i+(6i+1)(j-1)$

$i,j = 1,2,3,\ldots$

for primes in the sequence $N=6n+1$. Since all primes (except $2$ and $3$) are in one of two forms $6n−1$ or $6n+1$.

Can anybody advice me - is proposed "matrix sieve" algorithm well-known?

Answer 1

For the first proposition substitute $A=6x\pm 1$ and $B=6(x+y)\mp 1$, so that $A$ is an arbitrary positive integer coprime with $6$ and $B$ is any integer $\geq A$ with opposite sign modulo 6. For the second proposition do the same with $B=6(x+y)\pm 1$.

You are asking: is it true that $N$ coprime with 6 is composite if and only if it can be written as $N=AB$ with $A\leq B$?

Yes, it's obvious.