## Goldbach’s Conjecture over Z/mZ

Goldbach’s Conjecture states that every even integer greater than $2$ can be written as sum of $2$ primes. This problem is notoriously difficult, and still open.

However, one can prove the following version of Goldbach’s Conjecture for the ring $\mathbb{Z}/m\mathbb{Z}$:

Proposition. Let $m$ be a positive integer. Given any even integer $2n$ there exist primes $p$ and $q$ such that $2n=p+q \ (\textrm{mod } m)$.

The proof will, in fact, show that infinitely many such primes $p$ and $q$ exist. The required ingredients are Dirichlet’s Theorem on Arithmetic Progressions, and the following lemma due to Schinzel (1958):

Lemma. Given positive integers $m$ and $n$, there exists an integer $c$ such that $\textrm{gcd}(c, m) = \textrm{gcd}(2n-c, m)=1$.

Proof of Lemma. The case $m=1$ is trivial, so assume $m>1$ and write its prime factorization. We have $m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{r}^{\alpha_{r}}$ where $p_{1}, ... p_{r}$ are distinct primes, $\alpha_{1}, ... \alpha_{2}$ are some positive integers. For each $i\in\{1, 2, ... m\}$, we will show that there exists an integer $c_{i}$ such that $\textrm{gcd}(c_i, p_i) = \textrm{gcd}(2n-c_i, p_i)=1$. If $p_i=2$, we can let $c_i=1$. If $p_{i}$ is an odd prime, $p_{i}$ cannot divide both $2n-1$ and $2n+1$ (otherwise, it would divide their difference which is $2$); so if $\textrm{gcd}(2n-1, p_i)=1$ let $c_i=1$, and if $\textrm{gcd}(2n+1, p_{i})=1$ let $c_i=-1$. Now, by Chinese Remainder Theorem, one can find an integer $c$ such that $c\equiv c_i \ (\textrm{mod } p_{i})$ for each $i$. It is evident that  $\textrm{gcd}(c, m) = \textrm{gcd}(2n-c, m)=1$. (If it is not evident, assume $\textrm{gcd}(c, m)>1$. Then some prime divisor of $m$ would divide $c$. But it doesn’t.)

Proof of Proposition. From the lemma, we have $\textrm{gcd}(c, m) = \textrm{gcd}(2n-c, m)=1$ for some positive integer $c$. By Dirichlet’s Theorem, there exist primes $p$ and $q$ of the form $p=m k + c$ and $q=m \ell + (2n-c)$. Thus, $p + q = m(k+\ell)+2n$ which proves $2n=p+q \ (\textrm{mod } m)$, as desired. Indeed, Dirichlet’s Theorem guarantees the existence of infinitely many such primes $p$ and $q$.