Vinogradov Circle Method

Theorem
$\newcommand{\A}{\mathcal A}$Let $\A$ be a subset of the non-negative integers.

For $\alpha \in \R$, let $e(\alpha) = \exp\left({ 2 \pi i \alpha } \right)$.

Let


 * $\displaystyle T_N(s) = \sum_{\substack{a \in \A\\a \leq N}} s^a $

be the truncated generating function for $\A$.

Let $V_N(\alpha) := T_N(e(\alpha))$.

Let $r_{\A,\ell}(N)$ be the number of solutions $(x_1,\ldots,x_\ell) \in \A^\ell$ to the equation:


 * $\displaystyle x_1 + \cdots + x_l = N \qquad (1)$

Then:


 * $\displaystyle r_{\A,\ell}(N) = \int_0^1 V_N(\alpha)^\ell e(-N\alpha) \ d\alpha $

Proof
For $m \in \N$ let $r_{\A,\ell}(m; N)$ be the number of solutions to $(1)$ with no $x_i$ exceeding $N$.

Then $r_{\A,\ell}(m; N) = r_{\A,\ell}(m)$ for $m \leq N$, and $r_{\A,\ell}(m; N) = 0$ for $m > \ell N$.

Then we compute


 * $\displaystyle T_N(s)^\ell = \sum_{m = 0}^{\ell N} r_{\A,\ell}(m; N) s^m $

and


 * $\displaystyle V_N(\alpha)^\ell = \sum_{m = 0}^{\ell N} r_{\A,\ell}(m; N) e(\alpha m) \qquad (2)$

Now it follows from Exponentials Form Orthonormal Basis for $\mathcal L^2$ that


 * $\displaystyle \int_0^1 e(\alpha m) e(\alpha n)\ d\alpha = \delta_{mn}$

where $\delta_{mn}$ is the Kronecker delta. Therefore, we multiply $(2)$ by $e(-N\alpha)$ and integrate:


 * $\displaystyle r_{\A,\ell}(N) = r_{\A,\ell}(N; N) = \int_0^1 \displaystyle V_N(\alpha)^\ell e(-N\alpha)\ d\alpha $