# Questions about the proof that every odd integer is the sum of 5 primes

In http://arxiv.org/pdf/1201.6656.pdf, Tao proved that all odd numbers greater than 1 are the sum of 1, 3, or 5 primes. In page 36-37, he uses the fact that for all $x > 1.1\times10^{10}$, there exists some prime $x \le p \le x + x\cdot\frac{1}{2.8\times10^7}$ to show that if the binary Goldbach conjecture is verified up to $N_0$, then the ternary Goldbach conjecture is verified up to $(2.8\times10^7)N_0$, the 4 primes version is verified up to $(2.8\times10^7)^2 N_0$, and so on. How would one prove this?

Also, the author states that it suffices to show that Theorem 8.2 is true. However I do not understand how this is true. Can someone give me a proof of this statement?

Suppose that there is always a prime between $x$ and $(1+\epsilon)x$ for $x>C$ and that the Goldbach conjecture is verified for all even integers less than $N_0$ where $N_0>C$. Then take an arbitrary odd integer $N_0<n<\epsilon^{-1}N_0$. Given such an $n$, we may find a prime $p$ such that $$(1+\epsilon)^{-1}n\leq p\leq n.$$ Hence $$n-p\leq n-n(1+\epsilon)^{-1}\leq \epsilon n.$$ Since $\epsilon n\leq N_0$ by assumption, we may write $n-p$ as a sum of two primes. Hence all odd $n\leq \epsilon^{-1} N_0$ may be written as a sum of three primes. Applying this argument again, we find that all even numbers $n\leq \epsilon^{-2} N_0$ may be written as a sum of $4$ primes.
On the other hand, in  it is shown that every odd number larger than $\exp(3100)$ is the sum of three primes.