Vinogradov's Theorem

Theorem
Let $\Lambda$ be the von Mangoldt function.

For $N \in \Z$, let:


 * $\displaystyle \map R N = \sum_{n_1 + n_2 + n_3 \mathop = N} \map \Lambda {n_1} \, \map \Lambda {n_2} \, \map \Lambda {n_3}$

be a weighted count of the number of representations of $N$ as a sum of three prime powers.

Let $\mathcal S$ be the arithmetic function:


 * $\displaystyle \map {\mathcal S} N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$

where:
 * $p$ ranges over the primes
 * $p \nmid N$ denotes that $p$ is not a divisor of $N$
 * $p \divides N$ denotes that $p$ is a divisor of $N$.

Then for any $A > 0$ and sufficiently large odd integers $N$:


 * $\map R N = \dfrac 1 2 \map {\mathcal S} N N^2 + \map {\mathcal O} {\dfrac {N^2} {\paren {\log N}^A} }$

where $\mathcal O$ denotes big-O notation.

Proof of Theorem
Throughout the proof, for $\alpha \in \R$, let the following notation be understood:
 * $\map e \alpha := \map \exp {2 \pi i \alpha}$

Let $B > 0$, and set $Q = \paren {\log N}^B$.

For $1 \le q \le Q, 0 \le a \le q$ such that $\gcd \set {a, q} = 1$, let:


 * $\map {\mathcal M} {q, a} := \set {\alpha \in \closedint 0 1: \size {\alpha - \dfrac a q} \le \dfrac Q N}$

Let:
 * $\displaystyle \mathcal M := \bigcup {1 \mathop \le q \mathop \le Q} \bigcup_{\substack {0 \mathop \le a \mathop \le q} {\gcd \set {a, q} \mathop = 1} } \map {\mathcal M} {q, a}$

be referred to as the major arcs.

Let:
 * $\mathcal m := \closedint 0 1 \setminus \mathcal M$

be referred to as the minor arcs.

Lemma 1
By the Vinogradov Circle Method (with $\ell = 3$ and $\mathcal A$ the set of primes), letting $\displaystyle \map F \alpha = \sum_{n \mathop \le N} \map \Lambda n \, \map e {\alpha n}$ we have:


 * $\displaystyle \map R N = \int_0^1 \map F \alpha^3 \, \map e {-N \alpha} \rd \alpha$

So by splitting the closed unit interval into a disjoint union:
 * $\closedint 0 1 = \mathcal m \cup \mathcal M$

we have:


 * $\displaystyle \map R N = \int_{\mathcal m} \map F \alpha^3 \, \map e {-\alpha N} \rd \alpha + \int_{\mathcal M} \map F \alpha^3 \, \map e {-\alpha N} \rd \alpha$

We consider each of these integrals in turn.

Sum Over the Major Arcs
Putting these estimates together, we obtain:


 * $\map R N = \dfrac {N^2} 2 \map {\mathcal S} N + \map {\mathcal O} {\dfrac {N^2} {\paren {\log N}^{B / 2 - 5} } } + \map {\mathcal O} {\dfrac {N^2} {\paren {\log N}^{B/2} } }$

Now choose $B$ carefully.