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:


 * $\displaystyle \mathcal S_P \left({N}\right) := \sum_{q \mathop \le P} \frac {\mu \left({q}\right) c_q \left({N}\right)} {\phi \left({q}\right)^3}$
 * $\displaystyle \mathcal S \left({N}\right) := \lim_{P \to \infty} \mathcal S_P \left({N}\right)$

Then:
 * $\mathcal S \left({N}\right) = \mathcal S_P \left({N}\right) + \mathcal O(P^{\epsilon -1})$

and $\mathcal S$ has the Euler product:


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

where:
 * $p \mathop \nmid N$ denotes that $p$ is not a divisor of $N$


 * $p \mathop \backslash N$ denotes that $p$ is a divisor of $N$.

Proof
We have:

Trivially we have $\left\vert{\mu \left({q}\right)}\right\vert \le 1$.

Therefore:

Therefore by Convergence of Powers of Reciprocals:
 * $\mathcal S \left({N}\right) = \mathcal S_P \left({N}\right) + \mathcal O \left({P^{\epsilon - 1} }\right)$

as claimed.

By Euler Product, and because $\mu \left({p^k}\right) = 0$ for $k > 1$:


 * $\displaystyle \mathcal S \left({N}\right) = \prod_p \left\{ {1 + \frac {\mu \left({q}\right) c_q \left({N}\right)} {\phi \left({q}\right)^3} }\right\}$

Now for a prime $p$ we have:


 * $\mu \left({p}\right) = -1$


 * $\phi \left({p}\right) = p - 1$

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


 * $\displaystyle c_p \left({n}\right) = \sum_{d \mathop \backslash \gcd \left({p, n}\right)} \, \mathrm d \mu \left({\frac p d}\right)$

Let $p \mathrel \backslash N$.

Then: $\gcd \left({p, N}\right) = p$

which gives:


 * $c_p \left({N}\right) = p \mu \left({1}\right) + \mu \left({p}\right) = p - 1$

Let $p \nmid N$.

Then:
 * $\gcd \left({p, N}\right) = 1$

and so:
 * $c_p \left({N}\right) = -1$

Therefore:


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

This completes the proof.