Let $I(n)$ be the set of the first $n$ odd integers, $S_p(n)$ be a subset of the last $p$ elements of $I(n)$ such that $p$ is some prime number, and $a_{n-\left(\frac{p-1}{2}\right)}$ the $n-\left(\frac{p-1}{2}\right)_{th}$ element of $I(n)$.
Thinking on the Goldbach conjecture, I came to the following equivalence:
Goldbach conjecture is equivalent to affirm that every set $I(n)$ has some subset $S_p(n)$ such that $a_{n-\left(\frac{p-1}{2}\right)}$ is prime.
The proof is pretty straightforward. The sum of the elements of $I(n)$ is $n^2$, and the sum of the first $n-p$ elements of $I(n)$ is $\left(n-p\right)^2$. Therefore, the sum of the $p$ remaining elements is $n^2-\left(n-p\right)^2=-p^2+2np=p\left(2n-p\right)$.
Note that $a_{n-\left(\frac{p-1}{2}\right)}=2n-p$. Therefore, if both $p$ and $2n-p$ are prime numbers, Goldbach conjecture holds.
I am not sure that this approach adds any value for solving the conjecture. I share some of my thoughts:
Each prime $q_i$ generates $i$ subsets $S_i(n_i)$ such that $i$ is prime and $q_i=a_{n_i-\left(\frac{i-1}{2}\right)}$; so each prime $q_i$ guarantees that the conjecture holds for $i$ distinct values of $n$. Therefore, the first $i$ primes jointly guarantee that the conjecture holds for $1+2+3+...+i=\frac{î^2+i}{2}$ values of $n$. However, those values of $n$ are not necessarily distinct. The though part is proving that no value of $n$ is excluded.
Note also that some prime $q_i$ is the element $a_{\left(\frac{q-1}{2}\right)+1}$ of any set $I(n)$, and if $I(n)$ has some subset $S_p(n)$ such that $a_{n-\left(\frac{p-1}{2}\right)}$ is prime, then, noting that $a_{\left(\frac{q-1}{2}\right)+1}=a_{n-\left(\frac{p-1}{2}\right)}$, we have that $\left(\left(\frac{q-1}{2}\right)+1\right)+\left(\frac{p-1}{2}\right)=n$
Therefore, note that Goldbach conjecture would be equivalent to show that $\mathbb N-\{1,2\}\in P+Q$, where $P=\{x:\frac{p-1}{2}\}$, $Q=\{y:x+1\}$, and $P+Q=\{x+y:x\in P, y\in Q\}$. But this is kind of a tautology, as (I understand) does not add any information to solve the conjecture (if $p+q =2n$, then dividing by two $\frac{p}{2}+\frac{q}{2}=n$, and substracting $1$ from each numerator, and adding $1$, we get $\left(\left(\frac{q-1}{2}\right)+1\right)+\left(\frac{p-1}{2}\right)=n$.
I would like to know if this approach has something interesting, or does not add anything new to the problem.
Thanks!
Goldbach's conjecture postulates that every sufficiently large even number can be represented as the sum of two primes: $2n=p_1+p_2$. This is no different than saying that every sufficiently large integer is the average of two primes: $n=\frac{p_1+p_2}{2}$, which means $n$ lies equidistant between two primes, viz: $p_1+a=n=p_2-a$, or $p_1=n-a$ and $p_2=n+a$.
From this, we see $(n-a)(n+a)=p_1p_2=n^2-a^2$. But $a=n-p_1$ so $a^2=(n-p_1)^2$ and $n^2-a^2=n^2-(n-p_1)^2 =p_1(2n-p_1)$, which is just the formula OP arrived at in his statement of proof.
As I suggested in a comment, OP has rediscovered this known relationship in a clever new way.