I have developed the following conjecture:
Conjecture: For all twin prime pairs $(t_n, t_{n+1})$, $$12\mid t_n + t_{n+1}$$ such that $n > 1$.
but I am having trouble proving it. Can it be done?
My Attempt:
Since $t_{n+1} = t_n + 2$ then $$\begin{align} t_n +t_{n+1} &= 2(t_n +1) \\ \Leftrightarrow 6&\,\mid t_n + 1 \\ \Leftrightarrow t_n&\equiv -1\pmod 6.\end{align}$$ Lemma 1: Every prime $p > 5$ is congruent to $\pm 1\pmod 6$.
Proof: Let $n = 6q + r$ such that $q\in\mathbb{Z}^+$ and the remainder $r\in\{0,1,2,3,4,5\}$.
If $r = 0,2,4$ then $2\mid n$ and thus $n$ is not prime. If $r=3$ then $3\mid n$ and thus $n$ is not prime.
Therefore, if $n$ is prime, then $r\in\{1,5\}$. By letting $n\equiv \{-1,-5\}\pmod 6$ then this can be expressed as follows: $$\qquad \space \,\,\,n\equiv \pm1 \pmod 6.\tag*{$\Box$}$$
Let $\mathbb{P}_{>5}$ denote the set of primes $p > 5$, then since $n > 1$, we have that $t_n \in \mathbb{P}$ and thus so is $t_{n+1}$. $$\text{i.e.},\space\,\,\,\,\, t_n\equiv \pm 1\pmod 6.$$ Now if we let $t_n\equiv 1\pmod 6$ then for some $k\in\mathbb{N}$, we have that $t_{n+1} = 6k + 1$.
Lemma 2: $5$ is the only prime in two distinct twin prime pairs.
Proof: Every third odd number is divisible by $3$, thus no three successive odd numbers can be prime unless one of them is $3$. Therefore, $5$ is the only prime that is part of two twin prime pairs. $\qquad\,\,\,\,\,\Box$
Given the fact that $t_{n+1} = 6k + 1$ then if true, this also proves Lemma 2 since $t_{n+1} + 2 = 3(2k+1)$ and is hence not prime. Therefore, this case is perfectly plausible, but I want to prove otherwise.
How must I continue from here?
Thank you in advance.
Every pair $(m,m+2)$ of twin primes with $m>3$ has the form $(6n – 1, 6n + 1)$ for some $n$. The triple $(m,m+1,m+2)$ contains at least an even number, and only one multiple of $3$. Since $m$ and $m+2$ are both prime numbers, they cannot be even, so $m+1$ is even. Now, $m$ and $m+2$ cannot be a multiple of $3$. Indeed, otherwise, since they are prime we would have $m=3$, which is not possible, or $m+2=3$, that is $m=1$.
Hence $m+1$ is a multiple of $6$, hence the result. Now $(6n-1)+(6n+1)=12n$, showing that your conjecture is true.