Vinogradov's Theorem/Major Arcs/Lemma 1

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Lemma

Let $\phi$ be the Euler $\phi$ function.

Let $\mu$ be the Möbius function.

Let $c_q$ be the Ramanujan sum modulo $q$.


Let $P, N \ge 1$.



Let:

$\ds \map {\SS_P} N := \sum_{q \mathop \le P} \frac {\map \mu q \map {c_q} N} {\map \phi q^3}$
$\ds \map \SS N := \lim_{P \mathop \to \infty} \map {\SS_P} N$

Then:

$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon -1} }$

and $\SS$ has the Euler product:

$\ds \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$

where:

$p \nmid N$ denotes that $p$ is not a divisor of $N$
$p \divides N$ denotes that $p$ is a divisor of $N$.


Proof

We have:

\(\ds \size {\map {c_q} N}\) \(\le\) \(\ds \sum_{\substack {1 \mathop \le a \mathop \le q \\ \map \gcd {a, q} \mathop = 1} } 1\) Triangle Inequality
\(\ds \) \(=\) \(\ds \map \phi q\) Definition of Euler $\phi$ Function

Trivially we have $\size {\map \mu q} \le 1$.

Therefore:

\(\ds \size {\map \SS N - \map {\SS_P} N}\) \(=\) \(\ds \sum_{q \mathop > P} \frac {\size {\map \mu q \map {c_q} N} } {\map \phi q^3}\)
\(\ds \) \(\le\) \(\ds \sum_{q \mathop > P} \frac 1 {\map \phi q^2}\)
\(\ds \) \(\le\) \(\ds \sum_{q \mathop > P} q^{\epsilon - 2}\) for large $P$ and $\epsilon \in \R_{>0}$, by Asymptotic Growth of Euler Phi Function
\(\ds \) \(\le\) \(\ds P^{\delta - 1} \sum_{q \mathop > P} q^{\epsilon - \delta - 1}\) for any $\delta > \epsilon$

Therefore by Convergence of Powers of Reciprocals:

$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon - 1} }$

as claimed.

By definition of Euler product, and because $\map \mu {p^k} = 0$ for $k > 1$:

$\ds \map \SS N = \prod_p \paren {1 + \frac {\map \mu q \map {c_q} N} {\map \phi q^3} }$

Now for a prime $p$ we have:

$\map \mu p = -1$
$\map \phi p = p - 1$

We also have Kluyver's Formula for Ramanujan's Sum:

$\ds \map {c_p} n = \sum_{d \mathop \divides \map \gcd {p, n} } \rd \map \mu {\frac p d}$


Let $p \divides N$.

Then: $\map \gcd {p, N} = p$

which gives:

$\map {c_p} N = p \map \mu 1 + \map \mu p = p - 1$


Let $p \nmid N$.

Then:

$\map \gcd {p, N} = 1$

and so:

$\map {c_p} N = -1$

Therefore:

$\ds \map \SS N = \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} } \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} }$

This completes the proof.

$\blacksquare$