The corollary to the following theorem is "a mathematical expression by which we can know whether a given $(6m-1,6m+1)$ is a twin couple or not." I leave it to others to debate whether it falls outside of "general number sweeping."
Numbers of the form $6k \pm 1$, that is, all numbers having no factors of $2$ or $3$, form a semigroup closed under multiplication. Consequently, composite numbers in that semigroup have proper factors in that semigroup.
Consider composite numbers of the form $n=(6m-1)(6m+1)=36m^2-1$.
If $p_i \le 6m-1$ is a prime divisor of $n$, then $\exists s_j = \frac{n}{p_i} \ge 6m+1$. $p_i<s_j$ and $s_j$ has the form $6k \pm 1$
$n$ is a semiprime $\iff (6m-1,6m+1) \in \mathbb P \iff (6m-1,6m+1)$ are twin primes.
If $n$ is not a semiprime, it has at least one prime factor $p_i<6m-1\\$
Theorem: If $p_i=6a \pm 1 \le 6m-1 \land p_i \mid 36m^2-1$, then $m \equiv \pm a \bmod p_i$
Case 1: $p_i=6a-1 \Rightarrow s_j=\frac{36m^2-1}{p_i}=6(a+k)+1$.
$$36m^2-1=(6a-1)(6(a+k)+1) \\
36m^2-1=36a^2+36ak+6a-6a-6k-1 \\
36m^2=36a^2+36ak-6k$$
Since $36$ divides the LHS, $36$ must divide the RHS, so $k=6n,\ n \ge 0$. $\ n=0$ is formally allowed because $s_j=6a+1 > p_i=6a-1$, consistent with $p_i < s_j$. Substituting
$$m^2=a^2+6an-n \\
m^2-a^2=(m-a)(m+a)=n(6a-1)=n\cdot p_i \\
m \equiv \pm a \bmod p_i$$
Case 2: $p_i=6a+1 \Rightarrow s_j=\frac{36m^2-1}{p_i}=6(a+k)-1$.
$$36m^2-1=(6a+1)(6(a+k)-1) \\
36m^2-1=36a^2+36ak-6a+6a+6k-1 \\
36m^2=36a^2+36ak+6k$$
Since $36$ divides the LHS, $36$ must divide the RHS, so $k=6n,\ n > 0$. $n=0$ is not formally allowed, because in that case $s_j=6a-1 < p_i=6a+1$, which is inconsistent with $p_i < s_j$. Substituting
$$m^2=a^2+6an+n \\
m^2-a^2=(m-a)(m+a)=n(6a+1)=n\cdot p_i \\
m \equiv \pm a \bmod p_i$$
Corollary: $6m-1,6m+1 \in \mathbb P \iff \forall a<m,\ m\not \equiv \pm a \bmod p_i$.
Note that this corollary excludes the case $a=m$, for which $m\equiv a$ by any modulus. If $m=a$ (whence $n=0$ and $b=a$) is the only instance where $m \equiv \pm a \bmod p_i$, then $6m-1,6m+1$ are twin primes.
Note that this formulation does not resolve the twin prime conjecture. However, if the twin prime conjecture is false, then there exists some $m_0$ such that for all $m>m_0$, $m \equiv \pm a \bmod p_i$ for some $p_i=6a \pm 1 < 6m-1$.