Vinogradov's Theorem/Major Arcs/Lemma 1

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Lemma

Let $\phi$ be the Euler phi function, $\mu$ the Mobius function and $c_q$ the Ramanujan sum modulo $q$.

For $P,N \geq 1$, define:

$\displaystyle \mathfrak S_P(N) = \sum_{q \leq P}\frac{\mu(q)c_q(N)}{\phi(q)^3},\quad \mathfrak S(N) = \lim_{P \to \infty} \mathfrak S_P(N)$

Then $\mathfrak S(N) = \mathfrak S_P(N) + \mathcal O(P^{\epsilon -1})$ and $\mathfrak S$ has the Euler product:

$\displaystyle\mathfrak S(N) = \prod_{p \nmid N} \left( 1 + \frac 1{(p-1)^3} \right)\prod_{p \mid N} \left( 1 - \frac 1{(p-1)^2} \right)$

Firstly, we have

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \vert c_q(n) \vert$	$\leq$	$\displaystyle \sum_{\substack{1 \leq a \leq q\\\gcd(a,q) = 1} }1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	By the Triangle Inequality
	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \phi(q)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	By the definition of Euler's function

Trivially we have $|\mu(q)| \leq 1$. Therefore

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \vert \mathfrak S(N) - \mathfrak S_P(N) \vert$	$=$	$\displaystyle \sum_{q > P}\frac{\vert \mu(q)c_q(N) \vert}{ \phi(q)^3}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\leq$	$\displaystyle \sum_{q > P}\frac{1}{ \phi(q)^2}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\leq$	$\displaystyle \sum_{q > P} q^{\epsilon - 2}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	For large $P$ and any $\epsilon > 0$, by Asymptotic Growth of Euler Phi Function
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\leq$	$\displaystyle P^{\delta - 1}\sum_{q > P} q^{\epsilon-\delta-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	For any $\delta > \epsilon$

Therefore by Convergence of Powers of Reciprocals $\mathfrak S(N) = \mathfrak S_P(N) + \mathcal O(P^{\epsilon -1})$ as claimed.

By Euler Product, and because $\mu(p^k) = 0$ for $k > 1$, we have:

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

Now for a prime $p$ we have:

$\mu(p) = -1$

$\phi(p) = p - 1$

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

If $p \mid N$ then $\gcd(p,N) = p$, and this gives

$c_p(N) = p\mu(1) + \mu(p) = p-1$

If $p \nmid N$ then $\gcd(p,N) = 1$, and $c_p(N) = -1$. Therefore

$\displaystyle \mathfrak S(N) = \prod_{p\mid N} \left\{ 1 - \frac{1}{(p-1)^2} \right\}\prod_{p\nmid N} \left\{ 1 + \frac{1}{(p-1)^3} \right\}$

This completes the proof.

$\blacksquare$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \vert c_q(n) \vert\)	\(\leq\)	\(\displaystyle \sum_{\substack{1 \leq a \leq q\\\gcd(a,q) = 1} }1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi(q)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the definition of Euler's function

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \vert \mathfrak S(N) - \mathfrak S_P(N) \vert\)	\(=\)	\(\displaystyle \sum_{q > P}\frac{\vert \mu(q)c_q(N) \vert}{ \phi(q)^3}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \sum_{q > P}\frac{1}{ \phi(q)^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \sum_{q > P} q^{\epsilon - 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	For large $P$ and any $\epsilon > 0$, by Asymptotic Growth of Euler Phi Function
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle P^{\delta - 1}\sum_{q > P} q^{\epsilon-\delta-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	For any $\delta > \epsilon$