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Suppose Goldbach conjecture true: $\forall{n\geq{9}}$ integer, $\exists{p_1\geq{3},p_2\geq{3}}$ primes for which $2n=p_1+p_2$.

But $\forall{m>1}$ integer, $\forall{p_1\geq{3},p_2\geq{3}}$ primes, $2(4m+1)\neq{2p_1+2p_2}$ and $\forall{k,k'\neq{0}}$ integers $2(4m+3)\neq{2p_1+8k+2p_2+8k'}$. Thus $$2(4m+3-4m-1)=4\neq{2p_1+8k+2p_2+8k'-2p_1-2p_2}=8(k+k')$$ Or $$2(4m+3+4m+1)=2(8m+4)\neq{2p_1+8k+2p_2+8k'+2p_1+2p_2}.$$ It means that $\exists{m>1}$ integer, $\forall{p_1\geq{3},p_2\geq{3}}$ primes $1\neq{2(k+k')}$ which is always verified. Or there exists $m,k,k'$ for which (I am not sure about the existence of $m$) $4m+2-2k-2k'=2(2m+1-k-k')\neq{p_1+p_2}$. It means that there exists an even which is never equal to the sum of two primes ! Is this reasoning correct ? I suppose there is a fault, where is it, please ? Thank you !

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You say: for all $m \gt 1$ and primes $p_1, p_2 \geq 3$, we have $2(4m + 1) \neq 2 p_1 + 2 p_2$ which is true, and that additionally, for all $k, k' \neq 0$, we have $2(4m+3) \neq 8(k + k') + 2(p_1 + p_2)$, which is also true since you can divide by $2$ and subtract $4(k+k')$ and we have that the lhs is odd and the rhs even.

But we cannot conclude that for all $k, k'$ your third sentence is true since in general $A \neq B$ and $C \neq D$ doesn't imply $A - C \neq B - D$, and similarly goes for $+$. In your third sentence though, the below happens to be true:

$$ 2(4m+3-4m-1)=4\neq{2p_1+8k+2p_2+8k'-2p_1-2p_2} $$

Let's look at the second part:

$$ 2(4m+3+4m+1)=2(8m+4)\neq{2p_1+8k+2p_2+8k'+2p_1+2p_2}. $$

It's true iff we cancel the $2$'s and that's true. So we have: $$ 8m+4 \neq {8(k+k')+2(p_1 + p_2)}. $$

That is what I don't think is true.

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