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 $N$,


 * $\displaystyle R(N) = \frac12 \mathfrak S(N)N^2 + \mathcal O\left( \frac {N^2} {(\log N)^A} \right)$

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:


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

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
For sufficiently large $N$ the major arcs are pairwise disjoint, and the minor arcs are non-empty.

Proof
Suppose that for some admissible $a_1/q_1 \neq a_2/q_2$ we have $\mathfrak M(q_1,a_1) \cap \mathfrak M(q_2,a_2) \neq \emptyset$.

Then using the definition of the major arcs, for $\alpha$ in the intersection we have:

and

This shows that $N \leq 2Q^3 = 2(\log N)^3$.

But by Polynomial Dominates Logarithm at Infinity, this is impossible for sufficiently large $N$, so the major arcs must be disjoint.

Since the major arcs are pairwise disjoint, closed intervals, by Cover of Interval By Closed Intervals is not Pairwise Disjoint it is not possible that $\mathfrak M = [0,1]$, so $\mathfrak m \neq \emptyset$.

Now by the Vinogradov circle method (with $\ell = 3$ and $\mathcal A$ the set of primes), letting $\displaystyleF(\alpha) = \sum_{n \leq N} \Lambda(n)e(\alpha n)$ 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 into Lemma 2, 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.