I'm not experienced with writing papers or proofs, so I hope you will me correct my work, or point out where I might be going wrong. At any rate, I'm happy to have this out of my head. Thank you for reading.
Abstract: The sieve of Eratosthenes, the Chinese Remainder Theorem and Euclid's proof of infinite primes combine to provide the tools for generating infinite primes, infinite twin primes and infinite Polignac prime pairs.
Congruences are represented as usual, $n \equiv r \pmod{d}$, or as a simple tuple, $(r,d)$, or as a group of tuples $g = \{(r_1,d_1), (r_2,d_2), \dots, (r_n,d_n)\}$.
The sieve of Eratosthenes provides that each prime, $p$, marks as composite every number, $n$, for which $n \equiv 0 \pmod{p}$. As the sieve runs, the smallest unmarked $n$ must be prime, $p_n=n$. The first number marked as composite by $p_n$ is $p_n^2$. All composites of $p_n \lt p_n^2$ having been marked as composite by some other prime $p_i \lt p_n$. Therefore, any unmarked $n$, $p_n \lt n \lt p_n^2$ must be prime.
The Chinese Remainder Theorem (CRT) given the remainders, $\{r_1,r_2,\dots,r_n\}$, of the division of an integer, $I$, by coprime integers, $\{d_1,d_2,\dots,d_n\}$, uniquely defines the remainder of the division of $I$ by the product of these integers.
$$\forall d_i \text{ pairwise coprime, } P_n=d_1d_2 \dots d_n\\ I \pmod{P_n} = CRT(\{(r_1,d_1), (r_2,d_2), \dots, (r_n,d_n)\})\\ \forall r_i \not \equiv 0 \pmod{d_i} \implies I \text{ is coprime}\\ \text{If any } r_i \equiv 0 \pmod{d_i} \implies I \equiv 0 \pmod{d_i} $$
Generating prime numbers uses CRT with a set of continguous known primes, $\{p_1,p_2,\dots,p_n\}$, as divisors, with remainders, $\{r_1,r_2,\dots,r_n\}$, such that $\forall r_i \not \equiv 0 \pmod{p_i}$.
$$\forall p_i \text{ continguous known primes}, P_n=p_1p_2 \dots p_n\\ \forall (r_i,p_i) \not\equiv 0 \pmod{d_i}\\ \\I \pmod{P_n} = CRT(\{(r_1,p_1), (r_2,p_2), \dots, (r_n,p_n)\}) \\ I \lt p_n^2 \implies I \text{ is prime}\\ I = \text{the smallest value } \gt 1 \implies I = p_{n+1}$$
Let $M_n$ be a set of prime number congruence groups, initially $M_2 = \{\{(1,2)\}\}$. When prime $p_{n+1}$ is added, the set is updated by a cartesian product with congruences for $p_{n+1}$, excluding $(0,p_{n+1})$, and CRT is performed for each congruence group in the updated set. $$M_{n+1} = M_n \otimes \{(1,p_{n+1}), (2,p_{n+1}), \dots, (p_{n+1}-1,p_{n+1})\}$$ The result can be depicted as a matrix $h$ rows high and $w$ columns wide. Here is the matrix, $M_5$ for $p_{n+1}=5$; $$ h = \prod_{i=1}^n (p_i-1) \qquad w = (p_{n+1} - 1)\\ p_n = 3, \quad P_n = 6, \quad p_{n+1}=5, \quad P_{n+1} = 30\\ \begin{array} {cc|cccccc} \hline M_5, \pmb{\underline{T_5}} & \space & (1,5) & (2,5) & (3,5) & (4,5) & \text{ skipped} \\ \hline (1,2) & (1,3) & 1 & 7 & 13 & 19 & x=4, 25 = 5(5) \\ \space & (2,3) & \pmb{\underline{11}} & \pmb{\underline{17}} & 23 & \pmb{\underline{29}} & x=0, 5 = p_{n+1} \\ \end{array} \\ \space \\ $$
Proof of infinite prime numbers
$\bullet$ The smallest value, $s$, in every row of matrix $M_{n+1}$ is a value from matrix $M_n$, unless that value is $\equiv 0 \pmod{p_{n+1}}$. By CRT, such values cannot exist in $M_{n+1}$ and are replaced by $s+P_n$.
$\bullet$ Every value, $v$, in every row, although computed by CRT, can also be calculated as $v = s + P_nx$, where $x \in [0,1,\dots,p_{n+1}-1]$, except where $v \equiv 0 \pmod{p_{n+1}}$. By CRT, such values cannot exist in $M_{n+1}$. Every row of $M_{n+1}$ skips one value of $x$.
$\bullet$ Because the values in $M_{n+1}$ are scaled by $P_n$, the values of $M_{n+1}$ are consistently spread across $[1,\dots,p_{n+1}^2,\dots,P_{n+1}-1]$.
$\bullet$ One congruence group will have $\forall r_i=1$. CRT applied to this group produces $1 \pmod{P_{n+1}} = \{1,P_{n+1}+1,\dots\}$
$\bullet$ By Euclid's proof of infinite primes, $P_{n+1}+1$ is prime.
$\bullet$ By Eratosthenes and CRT, all values $v \in (1,\dots,p_{n+1}^2)$ are prime, and the second smallest value is $p_{n+2}$, which becomes $p_{n+1}$ for the next iteration.
$\bullet$ By Eratosthenes and CRT, all values $v \in (p_{n+1}^2,\dots,P_{n+1}-1)$ are prime or transient composites.
$\bullet$ Transient composites may appear, but cannot exist after their smallest prime factor is applied as $p_{n+1}$. For example; $$\begin{array} {llll} 209 &= CRT(\{(1,2), (2,3), (4,5), (6,7)\}) & \gt 7^2 = 11(19) \\ 209 &= CRT(\{(1,2), (2,3), (4,5), (6,7), (0,11)\}) & (0,p_i) \text{ Not allowed.} \end{array}$$ $\bullet$ Therefore, as $n \rightarrow \infty$, this method produces an infinite number of prime numbers and only prime numbers.
Generating twin primes: The twin prime conjecture states: there are infinitely many primes $p$ such that $p + 2$ is also prime. Consider a prime, $p_{n+1}$, constructed by the method defined above. Generating a twin prime is achieved by adding $2$ to each $r_i$ plus an additional congruence, $(2,p_{n+1})$; $$\begin{array} {llll} p_{n+1} &= CRT(\{(r_1,p_1), (r_2,p_2), \dots, (r_n,p_n)\})\\ I &= CRT(\{(r_1+2,p_1), (r_2+2,p_2), \dots, (r_n+2,p_n), (2,p_{n+1})\})\\ & \forall (r_i+2) \not \equiv 0 \pmod{p_i} \implies I = p_{n+2} \text{ twin prime of } p_{n+1} \end{array}$$
Let $T_n$ be a set of twin prime number congruence groups, initially $T_2 = \{\{(1,2)\}\}$. When prime $p_{n+1}$ is added, the set is updated by a cartesian product with congruences for $p_{n+1}$, excluding $(0,p_{n+1})$, and $(p_{n+1}-2,p_{n+1})$, and CRT is performed for each congruence group in the updated set. $$T_{n+1} = T_n \otimes \{(1,p_{n+1}), \dots (p_{n+1}-3,p_{n+1}), (p_{n+1}-1,p_{n+1})\}$$ The result can be depicted as a matrix, $T_{n+1}$, which is $h$ rows high and $w$ columns wide. $$ h = \prod_{i=1}^n (p_i-2) \qquad w = (p_{n+1} - 2)$$
Proof of infinite twin primes
$\bullet$ Because the congruences used to produce $T_{n+1}$ were used to produce $M_{n+1}$, less congruences $(p_{n+1}-2,p_i)$, $T_{n+1} \subset M_{n+1}$. The matrix $T_5 \subset M_5$, is shown above with twins values $\pmb{\underline{bold-underline}}$.
$\bullet$ The proof of infinite prime numbers for matrices $M_{n+1}$ applies to matrices $T_{n+1} \subset M_{n+1}$.
$\bullet$ By CRT, exclusion of all congruences $(p_i-2,p_i) \forall p_i$, this method can produce no values $p_{n+1}+2 \equiv 0 \pmod{p_i} \forall p_i$.
$\bullet$ Therefore, as $n \rightarrow \infty$, this method produces an infinite number of prime numbers and only prime numbers, and for each such prime $p_n+2$ is also prime.
Generating Polignac prime pairs: Polignac's conjecture states: for every natural number $k$, there are infinitely many primes $p$ such that $p + 2k$ is also prime. Consider a prime, $p_{n+1}$, constructed by the method defined above. Generating a Polignac twin prime is identical to generating a twin prime pair, but replacing the $2$'s with $2k$'s; $$\begin{array} {llll} p_{n+1} &= CRT(\{(r_1,p_1), (r_2,p_2), \dots, (r_n,p_n)\})\\ I &= CRT(\{(r_1+2k,p_1), (r_2+2k,p_2), \dots, (r_n+2k,p_n), (2k,p_{n+1})\})\\ & \forall (r_i+2k) \not \equiv 0 \pmod{p_i} \text{ and } I < p_{n+1}^2 \implies I\text{ is the Polignac twin of } p_{n+1} \end{array}$$
Let $T_n$ be a set of Polignac twin prime number congruence groups, initially $T_2 = \{\{(1,2)\}\}$. When prime $p_{n+1}$ is added, the set is updated by a cartesian product with congruences for $p_{n+1}$, excluding $(0,p_{n+1})$, and $(p_{n+1}-2k,p_{n+1})$, and CRT is performed for each congruence group in the updated set.
$$\begin{array} {llll}
T_{n+1} = T_n \otimes \{ & (1,p_{n+1}), \dots, (p_{n+1}-2k-1,p_{n+1}),\\ \space & (p_{n+1}-2k+1,p_{n+1}),\dots,(p_{n+1}-1,p_{n+1})\}
\end{array}$$
The result can be depicted as a matrix, $T_{n+1}$, which is $h$ rows high and $w$ columns wide.
$$ h = \prod_{i=1}^n (p_i-2) \qquad w = (p_{n+1} - 2)$$
Proof of infinite Polignac twin primes
$\bullet$ Because the congruences used to produce $T_{n+1}$ were used to produce $M_{n+1}$, less congruences $(p_{n+1}-2k,p_i)$, $T_{n+1} \subset M_{n+1}$. The matrix $T_5 \subset M_5$, is shown above with twin primes, $k=1$, values $\pmb{\underline{bold-underline}}$.
$\bullet$ The proof of infinite prime numbers for matrices $M_{n+1}$ applies to matrices $T_{n+1} \subset M_{n+1}$.
$\bullet$ By CRT, exclusion of all congruences $(p_i-2k,p_i) \forall p_i$, this method can produce no values $p_{n+1}+2k \equiv 0 \pmod{p_i} \forall p_i$ provided $p_{n+1}+2k \lt p_{n+1}^2$, or $$p_{n+1} \ge \lfloor \frac{1}{2}(1+\sqrt{8k+1}) \rfloor$$
$\bullet$ Therefore, as $n \rightarrow \infty$, this method produces an infinite number of prime numbers and only prime numbers, and for each such prime $p_n+2k$ is also prime.
