Prove that if every even natural number greater than 2 is a sum of two primes (the Goldbach Conjecture), then every odd natural number greater than 5 is a sum of three primes.

This is equivalent to asking if every even natural number greater than 2 is a sum of two primes,

\[ \begin{aligned} \forall{(n \in \mathbb{N} \land n > 1)} \exists{(p,q \in \mathbb{P})}[p + q & = 2n] \end{aligned} \]

is it true that:

\[ \begin{aligned} \forall{(m \in \mathbb{N} \land m > 2)} \exists{(r,s,t \in \mathbb{P})}[r + s + t & = 2m + 1] \end{aligned} \]

Exploiting m > 2 and n > 1:

\[ \begin{aligned} \forall{(m \in \mathbb{N} \land m > 2)} \exists{(x \in \mathbb{N})}[x & = m - 1 > 1] \\ m & = x + 1 > 2 \end{aligned} \]

Now,

\[ \begin{aligned} 2m + 1 & = 2(x + 1) + 1 \\ & = 2x + 3 \\ \end{aligned} \]

But x is greater than 1 which means 2x is an even number greater than 2, and therefore is a sum of two primes. Since 3 is prime, (2m + 1), an arbitrary odd natural number greater than 5, is therefore a sum of three primes.

# posted by rot13(Unafba Pune) @ 8:30 AM