# Why elements of the set can be Goldbach pairs for a given even number?

Let's take even number $100$ as an example (an example in the paper):

Fom $2$ to $\sqrt{100}$ there's four primes:$\ 2,\ 3,\ 5,\ 7.\$Let \begin{align*} &A=\{n: n \in \mathbb{Z^+}, n<2,n \neq 100 \mod 2\}=\{1\}\\ &B=\{n: n \in \mathbb{Z^+}, n<3,n \neq 100 \mod 3\}=\{2\}\\ &C=\{n: n \in \mathbb{Z^+}, n<5,n \neq 100 \mod 5\}=\{1,2,3,4\}\\ &D=\{n: n \in \mathbb{Z^+}, n<7,n \neq 100 \mod 7\}=\{1,3,4,5,6\}\\ &E=\{n: n \in \mathbb{Z^+}, n<100,n \mod 2 \in A,n \mod 3 \in B,n \mod 5 \in C,n \mod 7 \in D\}=\{11, 17, 29, 41, 47, 53, 59, 71, 83, 89\} \end{align*} We see that elements of the set $E$: $\ 11+89=100,\ 17+83=100, \ 29+71=100,\ 41+59=100, \,47+53=100$

Why elements of the set $E$ can be Goldbach pairs for the given even number 100?

Maybe there's an answer in that paper, but it may take pages and in bad math, hope there's an answer of coolness.Any revise,modify,suggestion,edit are welcomed.

• Vixra, not using Latex, and only one reference; any one of these alone is a bad sign. – user61527 Jul 13 '14 at 2:26
• ... Apply water to the burned area – Jorge Fernández Hidalgo Jul 13 '14 at 2:29
• @T.Bongers Thanks, Sage and Pari code for the example see:mymathforum.com/number-theory/… – miket Jul 13 '14 at 2:30
• Microsoft Word, no less. I'm not familiar with vixra; is it supposed to be a version of the arxiv for people ("for the whole community") who have no academic connections and thus can't get an arxiv account? – anomaly Jul 13 '14 at 3:07
• The ten elements of $E$ are the numbers that are 5 modulo 6, leaving out 5, 23, 35, 65, 77, and 95 --- which also add up to 100 in pairs (but without benefit of primality). – Gerry Myerson Jul 13 '14 at 3:54

Fix some large even $n$. The paper constructs the set $$A_p = \{0 < m \leq n\,|\, m\not\equiv 0, n\!\pmod{p}\}$$ for each prime $p\leq \sqrt{n}$. Now, if $1 < m < n$, then $m$ is certainly prime if $m\in A_p$ for all $p\leq \sqrt{n}$. We want to find such $m, m'$ with $m + m' = n$. Long story short, the paper then uses the Chinese remainder theorem to show that $$A = \bigcap_{p\leq \sqrt{n}} A_p$$ is nonempty. (Note that $n$ is even, so $A_2$ is nonempty.) That doesn't follow. There's certainly an element $d$ that satisfies the given system of congruences, but there's no reason why it should be on the interval $[0, n]$. (It's only determined mod $N = \prod_{p\leq \sqrt{n}} p$, which is generally much larger. Much larger, in fact; for $n = 10^4$, the product is more than $10^{36}$.) Without that criterion, it doesn't follow that $d$ is prime.