The twin prime conjecture posits that there are an infinite number of twin primes, or equivalently that there is no largest twin prime pair. I conjecture more specifically that for any twin prime pair $(p_i,p_{i+1})=(6m-1),(6m+1)$, the open interval $(p_i^2,\dots,p_{i+1}^2)=(36m^2-12m+1,\dots,36m^2+12m+1)$ contains at least one twin prime.
If my conjecture is true, it would prove the twin prime conjecture inductively. I have empirically verified my conjecture up to the twin prime pair $1091,1093$; the intervals between the squares of those primes, and between the squares of all smaller twin primes, contain twin primes. Moreover, the number of twin primes in the intervals tends to increase as the magnitude of the twin prime pair whose squares bound the interval increases, reaching several dozen in some cases. However, empirical observations and trends do not constitute a proof; they merely rule out readily discoverable counterexamples.
This post is lengthy, but it contains only four main components. FIRST, a Lemma. I have presented this idea in other posts, but I present it here so that readers will not have to look elsewhere to appreciate it. SECOND, the choice of particular intervals within which to look for twin primes. THIRD, boundary conditions which arise as a consequence of the Lemma and the chosen intervals. FOURTH, some practical illustration of how the ideas play out in a very few instances. However, the mathematics involved in moving from my empirical illustrations to proving a general theorem are above my level. Therefore, my question is: Does the existence of the boundary conditions I describe with respect to the intervals I examine permit progress toward proving the twin prime conjecture?
Definitions: Unless otherwise stated, in this discussion "prime" or "prime number" refers only to those natural numbers that are prime AND have the form $6k\pm 1$; i.e. primes $\ge 5$. $p_n$ denotes the $n$th such prime number, and $a_n$ denotes the natural number associated with $p_n$ such that $p_n=6a_n \pm 1$. A listing of $a_n$ can be found in OEIS A024699.
Lemma: $(6m-1),(6m+1)$ is a twin prime pair if and only if $\forall a_i<m,\ m\not \equiv \pm a_i \bmod p_i$
Proof: $m>a_i \iff 6m-1 >p_i$
$m\not \equiv \pm a_i \bmod p_i \iff m^2\not \equiv a_i^2 \bmod p_i \iff 36m^2-1\not \equiv 36a_i^2-1 \bmod p_i$ The final equivalence is true for all $p_i$ because $36$ is coprime to every $p_i$
$36m^2-1\not \equiv 36a_i^2-1 \bmod p_i \iff (6m-1)(6m+1) \not \equiv (6a_i-1)(6a_i+1) \bmod p_i \equiv 0 \bmod p_i$. The final equivalence is true by the definition of $a_i$
Thus, $\forall a_i<m,\ m\not \equiv \pm a_i \bmod p_i \iff (6m-1)(6m+1) \not \equiv 0 \bmod p_i$; in other words, no prime smaller than $6m-1$ divides either of $(6m-1),(6m+1)$; whence each of $(6m-1),(6m+1)$ is a prime number, and they are a twin prime pair. If $m\equiv \pm a_i \bmod p_i$ then one, but not both, of $(6m-1),(6m+1)$ is divisible by $p_i$. A listing of values of $m$ which meet these criteria, and give rise to twin primes, can be found in OEIS A002822. A further observation: Although the Lemma presumes testing with all $p_i< 6m-1$, as a practical matter one would only need to test with $p_i\le \sqrt{6m+1}$, because if a test with some $p_i> \sqrt{6m+1}$ excluded some particular $m$, it is certain that some smaller $p_i$ would already have excluded that value of $m$.
The interval: Assume that $6m\pm 1$ is a twin prime pair. If we wanted to screen the open interval bounded by the squares of those primes, i.e. $(36m^2-12m+1,\dots,36m^2+12m+1)$, for twin primes, we could perform a two step procedure: In the first step, we would use an Eratosthenes-like sieve of all the numbers in the interval (in this instance, to include using the primes $2$ and $3$) to eliminate composite numbers in the interval. If we were to do so, we could screen using primes $\le 6m-1$, because the interval contains only one composite number, all of whose prime factors are $>6m-1$, which is the semiprime $36m^2-1$. All other composite numbers in the interval have at least one prime factor $<6m-1$. In the Eratosthenes-like sieve being considered, we must screen with the prime $6m-1$ because in a few instances the number $36m^2+1$ is prime, so the pair $36m^2\pm 1$ can only be excluded by testing with $6m-1$. In the second step, having identified primes in the interval, we could check to see whether there were any twin primes.
By resorting to the teaching of the Lemma, however, we can look for twin primes more directly. We need screen only those numbers within the interval that are multiples of $6$: $[36m^2-12m+6,36m^2-12m+12,\dots,36m^2+12m-6]$, and test the number which result from dividing those numbers by $6$: $[6m^2-2m+1,6m^2-2m+2,\dots,6m^2+2m-1]$. We can represent those $4m-1$ consecutive numbers as $6m^2-2m+t$ where $1\le t\le 4m-1$. It is sufficient to use the same set of primes that are sufficient to perform the Eratosthenes-like screen on the entire interval, as the Lemma tells us that any of the numbers to be tested, when multiplied by $6$ and incremented or decremented by $1$, will be adequately screened for primality by precisely that set of primes. The sole exception is the number $6m^2$, which may have to be screened using the prime $6m-1$. We can finesse this complication by including the additional limitation $t\ne 2m$. When this limitation is included, testing numbers $6m^2-2m+t$ with primes $p_i$ such that $a_i<m$ accomplishes everything that would be achieved by testing further with $p_j$ such that $m\le a_j <6m^2$.
In this light, my conjecture can be restated as: $$\forall a_i < m, m\not \equiv \pm a_i \bmod p_i \Rightarrow \exists t, 1\le t \le 4m-1 \land t\ne 2m, \text{ such that }\\ \forall a_i < m, (6m^2-2m+t) \not \equiv \pm a_i \bmod p_i$$
Boundary conditions: In screening for suitable values of $6m^2-2m+t$, the basis value $6m^2-2m$ is subject to boundary conditions. The first boundary condition is imposed by virtue of its form. For example, numbers of this form can only take on the values $0,3,4 \bmod 5$, and can only take on the values $0,1,4,6 \bmod 7$. Similar limitations occur with respect to higher prime moduli. The second boundary condition results from the fact that since $6m\pm 1$ is a twin prime, $m$ itself is subject to the limitations imposed by the Lemma. Although the form of $6m^2-2m$ admits of being $\equiv 4 \bmod 5$, this only occurs when $m\equiv 1 \bmod 5$, which is forbidden in the present case because $m\not \equiv \pm 1 \bmod 5$. So within the context of $6m\pm 1$ being a twin prime, $6m^2-2m \equiv 0,3 \bmod 5$. Similarly, $6m^2-2m\equiv 1 \bmod 7$ only occurs when $m\equiv -1 \bmod 7$, which is forbidden for like reasons, whence $6m^2-2m \equiv 0,4,6 \bmod 7$. A quick look at primes up to $37$ suggests that $6m^2-2m$ can take on $\frac{p_i+1}{2}$ values in consideration of its form only, and that limitations attendant upon $m$ eliminate one of those possibilities, leaving $\frac{p_i-1}{2}$ possible values. This is based on a very limited number of examples, but it would be of interest to know if the either of those relationships could be proved to be generally true.
Empirical application: How do these boundary conditions help us look for candidate numbers $x$ that give rise to twin primes $6x\pm 1$? Let's look at a few actual cases. Evaluating numbers of the form $6m^2-2m+t \bmod 5$, when $m\equiv 0,2 \Rightarrow 6m^2-2m \equiv 0 \bmod 5$, so in order to meet the requirements of the Lemma $t \not \equiv 1,4 \bmod 5$. In this case, $t \equiv 0,2,3 \bmod 5$ is permitted. On the other hand, when $m\equiv 3 \Rightarrow 6m^2-2m \equiv 3 \bmod 5$, in order to meet the requirements of the Lemma $t \not \equiv 1,3 \bmod 5$, which means that $t\equiv 0,2,4$ is permitted. With respect to $\bmod 5$, then, we see that $t$ can never be $\equiv 1$ and is always permitted to be $\equiv 0,2$. The admissiblity of $t \equiv -1,3$ depends particularly on $m \bmod 5$. We can perform the same kind of analysis for higher primes.
With respect to $\bmod 7$, for example, if $m\equiv 0,5$, then $t\equiv 1,6$ are disallowed and $t\equiv 0,2,3,4,5$ are permitted. If $m\equiv 4$, then $t\equiv 2,4$ are disallowed and $t\equiv 0,1,3,5,6$ are permitted. Finally, if $m\equiv 2,3$, then $t\equiv 0,2$ are disallowed and $t\equiv 1,3,4,5,6$ are permitted. Here, there are no values of $t \bmod 7$ that are absolutely disallowed, but $t \equiv 3,5 \bmod 7$ are always allowed.
Similar analyses show that $t\equiv 0,3,6,7,10 \bmod 11; t\equiv 0,9,10,12 \bmod 13; t\equiv 2,4,8,9,15 \bmod 17; t\equiv 1,5,7,11,17 \bmod 19$ are also always allowed. The point is not to determine every allowed equivalent value of $t$ for every prime modulus, but to suggest (subject to more rigorous proof) that the existence of boundary conditions results in a plurality of permissible congruences, which makes likely the discovery of a suitable values of $6m^2-2m+t$ in the designated range that will give rise to twin primes.
What next? Of course, if we start with a specific twin prime with known value of $m$, we could simply perform the necessary calculations to prove (or disprove) the conjecture for that particular case. But as $m$ becomes increasingly large, calculations with regard to each prime $\le 6m-1$ will quickly become impractical. Even more generally, the existence of permitted congruences for $t$ might give rise to collection of possibilities for applying the Chinese Remainder Theorem, but it is unclear how one would prove that acceptable solutions fell within the designated range. What I wonder is, can the general existence of this kind of the kind of boundary conditions described above be used to prove that one or more numbers of the form $6m^2-2m+t$, where $6m\pm 1$ is known to be a twin prime, and $t \le 4m-1$, give rise to a twin prime?
Added in response to lulu's comments: There is so much detail in the specifics of my post that the arc of the argument and question seems to get lost. For arbitrary consecutive primes, lulu (correctly) points out that we can't even prove that the interval between their squares contains a prime number, much less a twin prime. She makes the point that there are "no tools" to approach the question. For arbitrary primes $(6k_1 \pm 1)<(6k_2 \pm 1)$ we can make no generally true statement about their residues or the residues of $k_1,k_2$ with respect to any modulus. That limitation is traversed when we look at twin primes $6m\pm 1$. There are constraints on $m$ with regard to all prime moduli, and those constraints impose further restrictions on their squares when they bound an interval. The point of my question is to ask whether the imposed constraints can serve as a tool that is unavailable in the more general case.