Vinogradov's Theorem

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

For $N \in \Z$, let:


 * $\displaystyle R \left({N}\right) = \sum_{n_1 + n_2 + n_3 \mathop = N} \Lambda \left({n_1}\right) \Lambda \left({n_2}\right) \Lambda \left({n_3}\right)$

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:


 * $\displaystyle\mathcal S \left({N}\right) = \prod_{p \mathop \nmid N} \left({1 + \frac 1 {\left({p-1}\right)^3} }\right) \prod_{p \mathop \backslash N} \left({1 - \frac 1{\left({p-1}\right)^2} }\right)$

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

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


 * $\displaystyle R \left({N}\right) = \frac 1 2 \mathcal S \left({N}\right) N^2 + \mathcal O \left({\frac {N^2} {\left({\log N}\right)^A} }\right)$

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

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

Let $B > 0$, and set $Q = \left({\log N}\right)^B$.

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


 * $\displaystyle \mathcal M \left({q, a}\right) := \left\{{\alpha \in \left[{0 \,.\,.\, 1}\right] : \left|{\alpha - \frac a q }\right| \le \frac Q N}\right\}$

Let:
 * $\displaystyle \mathcal M := \bigcup{1 \le q \le Q} \bigcup_{\substack{0 \mathop \le a \mathop \le q} {\operatorname{gcd} \left({a, q}\right) \mathop = 1}} \mathcal M \left({q, a}\right)$

be referred to as the major arcs.

Let:
 * $\mathcal m := \left[{0 \,.\,.\, 1}\right] \setminus \mathcal M$

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 $\displaystyle F \left({\alpha}\right) = \sum_{n \mathop \le N} \Lambda \left({n}\right) e \left({\alpha n}\right)$ we have:


 * $\displaystyle R \left({N}\right) = \int_0^1 F \left({\alpha}\right)^3 e \left({-N \alpha}\right) \ d \alpha$

So by splitting the closed unit interval into a disjoint union:
 * $\left[{0 \,.\,.\, 1}\right] = \mathcal m \cup \mathcal M$

we have:


 * $\displaystyle R \left({N}\right) = \int_{\mathcal m} F \left({\alpha}\right)^3 e \left({-\alpha N}\right) \ d \alpha + \int_{\mathcal M} F \left({\alpha}\right)^3 e \left({-\alpha N}\right) \ d \alpha$

We consider each of these integrals in turn.

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


 * $\displaystyle R \left({N}\right) = \frac {N^2} 2 \mathcal S \left({N}\right) + \mathcal O \left( \frac{N^2} {\left({\log N}\right)^{B / 2 - 5}} \right) + \mathcal O \left({\frac{N^2} {\left({\log N}\right)^{B/2}}}\right)$

Now choose $B$ carefully.