  This paper will give both the necessary and sufficient conditions required to
find a counter-example to the Goldbach Conjecture by using an algebraic
approach where no knowledge of the gaps between prime numbers is needed. To
eliminate ambiguity the set of natural numbers, $\mathbb{N}$, will include zero
throughout this paper. It will be shown there exists a counter-example to the
Goldbach Conjecture, given by $2a$ where $a \in \mathbb{N}_{> 3}$, if and only
if for each prime $p_i < a$ there exists some unique $q_i, \alpha_i \in
\mathbb{N}$ where $a < q_i < 2a$ and $2a = q_i + p_i$ along with the condition
that $\prod_{i = 1}^{\pi(a)}q_i = \prod_{i = 1}^{\pi(a)}p_i^{\alpha_i}$. A
substitution of $q_i = 2a - p_i$ for each $q_i$ from each sum gives the product
relationship $\prod_{i = 1}^{\pi(a)}(2a - p_i) = \prod_{i =
1}^{\pi(a)}p_i^{\alpha_i}$. Therefore, if a counter-example exists to the
Goldbach Conjecture, then there exists a mapping $\mathcal{G}_-:\mathbb{C} \to
\mathbb{C}$ where $$\mathcal{G}_-(z) = \prod_{i = 1}^{\pi(a)}(z - p_i) -
\prod_{i = 1}^{\pi(a)}p_i^{\alpha_i}$$ and $\mathcal{G}_-(2a) = 0$. A proof of
the Goldbach Conjecture will be given utilizing Hensel's Lemma to show $2a$
must be of the form $2a = p_i^{\alpha_i} + p_i$ for all primes up to $a$ when
$a > 3$. However, this leads to contradiction since $2a < a\#$ for all $a > 4$.
A similar method will be employed to give the necessary and sufficient
conditions when an even number is not the difference of two primes with one
prime being less than that even number. These proofs will have implications for
proving the Polignac Conjecture.
