Statement regarding Goldbach's Conjecture?

Question

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 :

$$ |-1 -10| = 11$$

Answer 1

I'll make the $\lambda$s positive for simplicity.

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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).

Answer 2

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.

Answer 3

I am gonna provide a counterexample just to justify @fleablood's Note text in his answer.

Let $m$ be any positive integer with $m \geq 2$ and $r, s \in \mathbb{Z}^+$ such that $r \leq s$ and $2n = r + s$ with $r$ and $s$ not necessarily prime. We can prove that there exists at least one $\lambda_1$ and $\lambda_2$ such that

$$m = rq_1 + \lambda_1$$ $$m = sq_2 + \lambda_2$$ and $$|\lambda_1 + \lambda_2| \equiv 0 \pmod{r}$$ for some $\lambda_1$ and $\lambda_2$ being integers, but $r$ and $s$ are not even primes numbers.

Take $m = 13$, $r = 12$ and $s = 14$. So, $$26 = 2*13 = 12 + 14$$

$$13 = 12*1+1$$ and $$13 = 14*1 - 1$$

However, $|\lambda_1 + \lambda_2| \equiv |1 - 1| \equiv 0 \pmod{12} \equiv 0 \pmod{14}$ and for sure $r = 12$ and $s = 14$ are not prime numbers because they are even numbers. Also, notice that you said that "and $m$ is not prime" but in this counterexample it did work.