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Let $n \in \Bbb{N}, n \gt 1$. Let $\Bbb{P} = $ the prime numbers in $\Bbb{N}$. Define \begin{align*} A_n &= \{ (p,q) \in \Bbb{P}^2 : p + q = 2n\}, \\ B_n &= \{ (p, q) : p - q = 2n \}. \end{align*}

I conjecture the only way to reach higher sets is to union the partitions of lower sets in this way: $$A_n = \bigcup_{k=1}^{n-1} B_{n-k} \circ A_k,$$ where $\circ$ is relational composition.

This comes from the fact that if $(q, p) \in A_k$ and $(r, q) \in B_{n-k}$ then $(p, r) \in A_n$ necessarily.

Do you think this is true?

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You are asking whether the following is true for any integer $n>1$ and any primes $p,q$: \begin{multline*} p+q = 2n \iff \text{there exists an integer $1\le k\le n-1$ and a prime $\ell$ such that } \\ p-\ell = 2(n-k) \text{ and } \ell+q = 2k. \end{multline*} As you point out, the $\Longleftarrow$ direction is straightforward. The $\Longrightarrow$ direction is easy as long as $p\ge5$: just take $\ell=3$ and $k=(q+3)/2$, so that $p-\ell=(2n-q)-3=2n-2k$, for example. However, it fails when $p=3$, since there is no smaller odd prime to choose for $\ell$. (It also fails when $n=2$.)

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I'd argue no, because $$7+3=10\land7\cdot\frac{3}{2}+\frac{1}{2}=11\land3\cdot\frac{3}{2}+\frac{1}{2}=5 \implies 16=11+5$$via Collatz mapping: $$\frac{3n}{2}+1\land n=p+q\implies \frac{3n}{2}+1=(\frac{3p}{2}+\frac{1}{2})+(\frac{3q}{2}+\frac{1}{2})$$ . See https://oeis.org/A158709 , for relevant primes to pick from to add, that can if summed to 2 mod 4, turn 1 goldbach partition into another for the even just around the Collatz graph bend. it's technically hard to be truly composition free (okay technically, this is an iterated( repeated composition with self) map, but knowing it we need not iterate it) .

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