Statement regarding Goldbach's Conjecture? Question
I think using elementary (but twisted) means I can prove an interesting statement and was curious how a number theorist would prove the same. 
Let us we want to find $2$ primes which satisfy:
$$p_1 + p_2 = 2 m$$
and $m$ is not a prime. Then I can show:
$$ m = n_1 p_1 + \lambda_1$$
And 
$$ m = n_2 p_2 + \lambda_2$$
where $n$ and $\lambda$ are the negative remainders (see example).
Then:
$$  |(\lambda_1 +  \lambda_2)|  = \lambda p_1$$
where $\lambda$ is an integer (and $p_1$ is the smaller prime). The converse is true as well.
Example $1$
Consider $18$ which is the sum of $13$ and $5$
$$  13 + 5 = 2 \times 9 $$
Now, this can be expresses as as a negative remainder as:
$$ 9 = 13 \times 1 - 4 $$
$$ 9 = 5 \times 2 - 1$$
Verifying : 
$$ |-1 -4| = 5$$
Example $2$
Consider $24$ which is the sum of $13$ and $11$
$$  13 + 11 = 2 \times 12 $$
Now, this can be expresses as as a negative remainder as:
$$ 12 = 13 \times 1 - 1 $$
$$ 12 = 11 \times 2 - 10$$
Verifying : 
$$ |-1 -10| = 11$$
 A: I'll make the $\lambda$s positive for simplicity.

Since $p_1<p_2$ we have $$p_2>{p_1+p_2\over 2}=m$$ and so $n_2=1$. So $$p_1n_1+p_2n_2=p_1n_1+p_2=(p_1+p_2)+p_1(n_1-1)=2m+p_1(n_1-1).$$
Why is this relevant? Well, we also have by definition that $$2m=p_1n_1-\lambda_1+p_2n_2-\lambda_2$$ and so $$\lambda_1+\lambda_2=p_1n_1+p_2n_2-2m.$$
Combining these equalities we get $$\lambda_1+\lambda_2=2m+p_1(n_1-1)-2m=p_1(n_1-1).$$ So $p_1\vert \lambda_1+\lambda_2$ as desired (and in fact we've found the relevant multiple).
A: If $p_1 < p_2$ and $p_1 + p_2 = 2m$ then
$p_1 < m < p_2$ and $m-p_1= p_2 -m$
Lets let $d:= m-p_2 = p_2 -m$.
$p_2 > m$ so if $m=n_2p_2 + \lambda_2$ with $\lambda_2$ is the "negative remainder, then $n_1 = 1$ and $\lambda_2 = m-p_2=-d$.
$m = n_1p_1 + \lambda_1$ and $p_1 <m$ so $n_1\ge 2$ and  $\lambda_1 = m-n_1p_1=(p_1+d)-n_1p_1= d+(1-n_1)p_1$.
So $|\lambda_1 + \lambda_2|=|-d + d + (1-n_1)p_1| = (n_1 -1)p_1$.
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Note: this has nothing to do with primes and would be true if $p_1, p_2$ were not prime, and would also be true if $m$ were prime.
