Vinogradov's Theorem

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

For $N \in \Z$, let:


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

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

Let $\mathfrak S$ be the arithmetic function:


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

where $p$ ranges over the primes.

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


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

Outline of Proof
Goes here

Proof of Theorem
Throughout the proof, for $\alpha \in \R$, $e(\alpha) = \exp(2 \pi i \alpha)$.

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

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


 * $\displaystyle\mathfrak M(q,a) = \left\{ \alpha \in [0,1] : \left| \alpha - \frac aq \right| \leq \frac QN \right\}$

Moreover let:


 * $\displaystyle \mathfrak M = \bigcup{1 \leq q \leq Q} \bigcup_{\substack{0 \leq a \leq q}{\operatorname{gcd}(a,q) = 1}} \mathfrak M(q,a)$

the major arcs and $\mathfrak m = [0,1] \backslash \mathfrak M$, the minor arcs.

Lemma 1
Now by the Vinogradov circle method (with $\ell = 3$ and $\mathcal A$ the set of primes), letting $\displaystyle F(\alpha) = \sum_{n \leq N} \Lambda(n)e(\alpha n)$ we have:


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

So by splitting the unit interval into a disjoint union $[0,1] = \mathfrak m \cup \mathfrak M$ we have:


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

We consider each of these integrals in turn.

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


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

Now choose $B$ carefully.