Vinogradov's Theorem/Major Arcs/Lemma 1

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:

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

Therefore:

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.