The set of all sums of 3 primes covers almost all of $2\Bbb{Z}+1$ (result of Helfgott), what can one say about $\Bbb{P} - \Bbb{P} = 2\Bbb{Z}$ problem?

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Consider the set $\Bbb{P} = \pm$ the prime numbers and $\Bbb{P}_o$ is similarly $\pm$ the prime numbers other than $2$.

By Helfgott's result on the ternary Goldbach conjecture:

Every odd integer greater than $5$ is expressible as the sum of exactly 3 prime numbers (2 included here).

We also have this almost immediate corollary:

$$ 4\Bbb{P} = \Bbb{P} + \dots + \Bbb{P} = \{ p + q + r + s : p, q, r,s \in \Bbb{P}\} = \Bbb{Z} $$

So if you quotient $\Bbb{Z}$ setwise by the equivalence relation:

$$ (p, q, r,s)\sim (p', q', r',s') \iff p + q + r + s = p' + q' + r' + s' $$

You get the setwise isomorphism (bijection):

$$ \psi:\dfrac{\Bbb{P}^4}{\sim} \simeq \Bbb{Z}, \ \psi ([p,q,r,s]):= p + q +r+s $$

Now use a "transfer of structure" isomorphism technique:

$$ x,y \in \Bbb{P}^4/\sim \implies x * y := [\psi^{-1}(\psi(x) + \psi(y))] $$

to give the quotient of 4-tuples of $\pm$ prime numbers the structure of an anbelian group.

Then I wish to prove that $\Bbb{P}_o^2/\sim$ similarly defined but for $2$-tuples of odd primes is a subgroup, namely the one isomorphic with $2\Bbb{Z}$.


Clearly there is an embedding of sets:

$$ \iota:\Bbb{P}_o^2 \hookrightarrow \Bbb{P}^4 $$

Equivalence classe $[\cdot]$ on the left is defined similarly to be $[p,q] = [r,s] \iff p + q = r + s$. This ensures that $\iota$-induced morphism is indeed a group homomorphism once we quotient. Let:

$$ \iota([p,q]):= [p,q,r,-r] $$ for some fixed and arbitrary $r \in \Bbb{P}$, and $p, q \in \Bbb{P}_o$.

Then the group law on $\star$ on $\Bbb{P}_o^2/\sim$ is...

Well, make sure you know first that $\sim$ is defined as $(p,q) \sim (r,s) \iff p + q = r + s$ hence "similarly".

Then you define:

$$ h := [\iota(x)], h:\Bbb{P}_o^2 \to \Bbb{P}^4/\sim \\ \ \\ \ [p,q],[r,s]\in \Bbb{P}_o^2/\sim \implies \\ \ [p,q]\star[r,s]:= \\ \ [ h^{-1}(h(p,q) + h(r,s)) ] $$

This is indeed well defined because if $[p,q] = [p', q']$ then they sum the same so that they also sum the same in $\Bbb{P}^4$, so map to the same element $h(p,q) = h(p', q')$ and we're done.

It is a group homomorphism, by definition. Namely we deployed the rarely seen used technique of "transfer of structure" between an algebraic structure and a set.

Question. It seems too weird that I would have proven $\Bbb{P}_o - \Bbb{P}_o = \{p -q : p, q\in \Bbb{P}_o\} = 2\Bbb{Z}$ a well known open problem. So please be kind enough to point out the flaw since I was kind enough to present the proof in full on 2 sheets of paper.

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You define a binary operation '$\star$' on $\Bbb{P}_o^2$ and claim that this defines a group operation. But it is not clear that this binary operation is even well-defined; in writing $$[a,b]\star[c,d]:=[h^{-1}(h(a,b)+h(c,d))],$$ you seem to assume that $h^{-1}(p,q,r,s)$ exists for all $(p,q,r,s)\in\Bbb{P}^4/\sim$, or at least those tuples in the image of $\Bbb{P}_o^2$. Phrased differently, you assume that for all $n\in\Bbb{Z}$ there exists $p,q,r\in\Bbb{P}$ such that $n=p+q+r-r$, or at least for all even $n$. That is to say, you assume the Goldbach conjecture.