Goldbach's Conjecture — Direct & Contrapositive Proof

Authored by Khusraw (Ismail Mostafanejad) — 2–4 July 2026

Part I: Direct Proof

Theorem (Goldbach's Conjecture)

For all even integers $n > 2$, there exist prime numbers $p, q$ such that $n = p + q$.

Setup & Definitions

All primes $p > 2$ are odd (already proven by mathmaticians). Therefore:

\[P_{>2} = \{ n \in \mathbb{N} \mid n > 2 \text{ and the only divisors of } n \text{ are } 1, n \} \subset O = \{ n \in \mathbb{Z} \mid \exists k \in \mathbb{Z} : n = 2k + 1 \}\]

Note that $\mathbb{N}$ is a subset of $\mathbb{Z}$ and it can be substituted by $\mathbb{N}$ above, but by division I got even numbers from $\mathbb{Z}$.

Then it is satisfying if we prove it for $O \supset P_{>2}$. We use the dialectical methodology: Trails → Deductive → Inductive.

Empirical Trails

Step 1: $n = 4 \Rightarrow 1 + 3 = 4$ (1 and 3 are both primes)
Step 2: $n = 6 \Rightarrow 1 + 5 = 3 + 3 = 6$ (1, 3, 5 are all primes)
Step 3: $n = 8 \Rightarrow 1 + 7 = 3 + 5 = 8$ (1, 3, 5, 7 are all primes)
…
Step m: $n \Rightarrow p + q = n$
Step m+1: $n + 2 \Rightarrow p + q = n + 2$

Left-Side Proof (Even)

Assume $p = 2k + 1$ and $q = 2l + 1$ (odd primes):

\[p + q = (2k + 1) + (2l + 1) = 2k + 2l + 2 = 2(k + l + 1) = 2S \quad (\text{even})\]

Right-Side Proof (Even)

Assume $n = 2K$ is even at step $m$. Then at step $m+1$:

\[n + 2 = 2K + 2 = 2(K + 1) = 2S \quad (\text{even})\]

Valid for $P_{>2} \subset O$. Both left-side and right-side proofs of the equation are even (Note: we already knew that both sides are equal but only needed to proof even), therefore satisfying.

Part II: Contrapositive of Goldbach's Conjecture as a New Theorem

Theorem

For all integers $n$, if $n$ cannot be written as the sum of any two prime numbers, then $n$ is an odd integer or $n \le 2$.

Empirical Trails

Step 1: $n = 3 \Rightarrow 1 + 2 = 3$ (1 and 2 are primes — Exception)
Step 2: $n = 5 \Rightarrow 1 + 4 = 5$ (4 is not prime)
Step 3: $n = 7 \Rightarrow 3 + 4 = 7$ (4 is not prime)
…
Step m: $n \Rightarrow p + q = n$
Step m+1: $n + 2 \Rightarrow p + q = n + 2$

Left-Side Proof (Odd)

Assume $p = 2k + 1$ (odd) and $q = 2l$ (even):

\[p + q = 2k + 2l + 1 = 2(k + l) + 1 = 2S + 1 \quad (\text{odd})\]

Right-Side Proof (Odd)

Assume $n = 2K + 1$ (odd). Then at step $m+1$:

\[n + 2 = (2K + 1) + 2 = 2(K + 1) + 1 = 2S + 1 \quad (\text{odd})\]

Therefore, left-side and right-side proofs are both odd (Note: we already knew that both sides are equal but only needed to proof odd), therefore satisfying and valid for $n > 3$.

Academic Archival & Copyright License

DOI: 10.5281/zenodo.21682740