# Vinogradov's Theorem

## Contents

## 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.

### Corollary 1

Let:

- $\displaystyle r \left({N}\right) = \sum_{p_1 + p_2 + p_3 \mathop = N} 1$

where $p_1, p_2, p_3$ are prime.

Then for sufficiently large odd integers $N$:

- $r \left({N}\right) = \mathcal S \left({N}\right) \dfrac {N^2} {2 \left({\log N}\right)^3} \left({1 + \mathcal O \left({\dfrac{\log \log N} {\log N}}\right)}\right)$

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

### Corollary 2

Every sufficiently large odd integer is the sum of three prime numbers.

## Outline of Proof

## 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

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

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 Minor Arcs

For any $B > 0$:

- $\displaystyle \int_\mathcal M \map F \alpha^3 \map e {-\alpha N} \rd \alpha \ll \frac {N^2} {\paren {\ln N}^{B/2 - 5} }$

$\Box$

### Sum Over the Major Arcs

Let $B \in \R_{>0}$.

Then:

- $\displaystyle \int_\mathcal M \map F \alpha^3 \map e {-N \alpha} \rd \alpha = \frac {N^2} 2 \map {\mathcal S} N + \map {\mathcal O} {\frac {N^2} {\paren {\ln N}^{B/2} } }$

where the implied constant depends only on $B$.

$\Box$

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.

$\blacksquare$

## Source of Name

This entry was named for Ivan Matveevich Vinogradov.

## Sources

- 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Vinogradov's theorem**