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Am I wrong? Is my proof clear? Is this already a known result? If so, are there any simple implications or corollaries of this proof? (if that makes sense) I know some about groups, rings and fields, but not enough to argue with them.

Suppose $n,k\in\mathbb{N}$ such that $n=2k$, and $k>2$.
$\exists x,y\in\mathbb{N}$ such that $x+y=k-1$.
I claim that if $2x+1$ and $2y+1$ are prime then $n$ is the sum of two primes.

$$\begin{align*} n&=2k,\\ &=2(x+y+1),\\ &=4xy+2(x+y+1)-4xy,\\ &=4xy+2x+2y+2-4xy,\\ &=(4xy+2x+2y+1)+1-4xy,\\ &=(4xy+2x+2y+1)+1-((2x+1)-1)((2y+1)-1),\\ &=(2x+1)(2y+1)+1-((2x+1)-1)((2y+1)-1). \end{align*}$$

Let $m,l\in\mathbb{N}$ such that $m=2x+1$ and $l=2y+1$.

Then $$n=ml+1-(m-1)(l-1)$$ and

$ \forall p_{1},p_{2}\in\mathbb{N}$ such that $p_{1}$ and $p_{2}$ are prime, if $m=p_{1}$ and $l=p_{2}$
$$\begin{align*} n&=2k,\\ &=ml+1-(m-1)(l-1),\\ &=p_{1}p_{2}+1-(p_{1}-1)(p_{2}-1),\\ &\therefore n=p_{1}+p_{2} \end{align*}$$

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  • $\begingroup$ Welcome to MSE. Proof of what? $\endgroup$ Mar 7, 2021 at 7:12
  • $\begingroup$ The question shows some effort even if the result turned out to be trivial so I don't think downvotes are warranted. $\endgroup$
    – Oussema
    Mar 7, 2021 at 7:24
  • $\begingroup$ lol and thank you. I admit I don't really know what I'm talking about and that I am basically repeating myself. It seems like the way I arrived at the result is worth looking at though. I just wondered if anyone knew of any methods of proof that somehow utilized this, or about any results that seemed apparent from looking at the sum of two primes this way $\endgroup$
    – matt
    Mar 7, 2021 at 7:39
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    $\begingroup$ "I just wondered if anyone knew of any methods of proof that somehow utilized this" You mean algebraic manipulation? If so then it has been utilized extensively. $\endgroup$
    – Oussema
    Mar 7, 2021 at 7:55
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    $\begingroup$ That's a roundabout derivation of $\ 2(x+y+1) = 2x\!+\!1\,+\, 2y\!+\!1\ \ $ $\endgroup$ Mar 7, 2021 at 8:26

1 Answer 1

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I'm afraid the result is quite trivial: $$x+y=k−1$$ $$2x+2y=2k-2=n-2$$ $$(2x+1)+(2y+1)=n$$ So if $2x+1$ and $2y+1$ are prime then $n$ is the sum of two primes.

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  • $\begingroup$ "I'm afraid the result is quite trivial" yeah, but in a weird way. I'm asking more if anyone could expound some about the specific criteria I start with (n=2k, x+y=k−1) and the way I arrived at that trivial conclusion than I am about the fact. $\endgroup$
    – matt
    Mar 7, 2021 at 7:33
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    $\begingroup$ What is there to say about starting with something simple and then finding a complicated way to do something simple with it? $\endgroup$ Mar 7, 2021 at 8:59
  • $\begingroup$ "What is there to say about starting with something simple and then finding a complicated way to do something simple with it?" Isn't the whole idea of Abstract Algebra to start with something simple and find a complicated way to do something else simple with it? $\endgroup$
    – matt
    Mar 7, 2021 at 14:08
  • $\begingroup$ Let me know, Matt, when you have found a simple way to prove that there's no formula for solving fifth degree equations in radicals. $\endgroup$ Mar 7, 2021 at 22:08

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