This note discusses the idea of efficient congruencing related to Vinogradov's mean value theorem. This method is really similar to $p$–adic concepts but replaces them with Archimedean counterparts. Also, both papers contain an iterative analysis. This is due to the original congruence modulo $p^a$ not being that powerful; we raise the power of $p$ and repeat the lifting process; hence, it is an iterative process.
The problem we are concerned with here is a mean value that also counts the number of solutions to the system of equations or congruences. As in the typical Vinogradov's mean value theorem, let $k\in\mathbb{N}$ and $\boldsymbol{\alpha}\in\R^k$; we consider the mean value \[ J_{s,k}(X) \;=\; \int_{[0,1)^k} \bigl|f_k(\boldsymbol{\alpha};X)\bigr|^{2s}\, d\boldsymbol{\alpha}, \] where \[ f_k(\boldsymbol{\alpha};X) = \sum_{1\le x\le X} \e\bigl(\psi(x;\boldsymbol{\alpha})\bigr),\qquad \psi(x;\boldsymbol{\alpha}) = \alpha_1 x + \cdots + \alpha_k x^k. \]
It follows from the orthogonality that, for $s\in\mathbb{N}$, this mean value counts the number of integral solutions to the system of equations \[ x_1^j + \cdots + x_s^j \;=\; y_1^j + \cdots + y_s^j \qquad (1\le j\le k), \] with $1\le x_i,y_i \le X$.
For $f_k(\boldsymbol{\alpha};X)$, we sort the summation into arithmetic progressions modulo $p^c$, $c$ non–negative integers, and by Hölder's inequality, we now look at \[ f_c(\boldsymbol{\alpha};\boldsymbol{\xi}) \;=\; \sum_{\substack{1\le x\le X\\x\equiv \xi \,(\bmod p^c)}} \e\bigl(\psi(x;\boldsymbol{\alpha})\bigr). \] For brevity, we write \[ \int_{[0,1)^k} f(\boldsymbol{\alpha}) \, d\boldsymbol{\alpha} \;=\; \oint f(\boldsymbol{\alpha}) \, d\boldsymbol{\alpha}. \]
We first look at the typical treatment by analyzing the following mean value \[ \oint \bigl|f(\boldsymbol{\alpha};X)\bigr|^{2k} \bigl|f_1(\boldsymbol{\alpha};\boldsymbol{\xi})\bigr|^{2s} \, d\boldsymbol{\alpha}, \] which counts the number of integral solutions of the system \[ \sum_{i=1}^k (x_i^j - y_i^j) = \sum_{\ell=1}^s \bigl((p u_\ell + \xi)^j - (p v_\ell + \xi)^j\bigr) \qquad (1\le j\le k), \] with $1\le x_i,y_i\le X$ and $0\le u_\ell,v_\ell\le (X-\xi)/p$.
By writing $x = (x + \xi)-\xi, \text{and } y = (y + \xi)-\xi,$ and taking a binomial expansion of this system, \[ \sum_{i=1}^k \bigl( (x_i-\xi)^j - (y_i-\xi)^j \bigr) = p^j \sum_{\ell=1}^s \bigl( u_\ell^j - v_\ell^j \bigr) \qquad (1\le j\le k), \] whence \[ \sum_{i=1}^k (x_i-\xi)^j \equiv \sum_{i=1}^k (y_i-\xi)^j \pmod{p^j} \qquad (1\le j\le k). \]
This above congruence provides the efficient congruence condition, and by Hensel's lemma, for a well–conditioned set of $\mathbf{x}$, we have $\mathbf{n}\in\Z^k$ satisfying \[ \sum_{i=1}^k (x_i-\xi)^j \equiv n_j \pmod{p} \qquad (1\le j\le k). \] Then we have such $\mathbf{x}$ as a unique solution satisfying \[ \sum_{i=1}^k (x_i-\xi)^j \equiv n_j \pmod{p^k} \qquad (1\le j\le k). \]
From the efficient congruence condition, we deduce that \[ \mathbf{x} \equiv \boldsymbol{y} \pmod{p^k}, \] with a multiplicity $k!\,p^{\frac12 k(k-1)}$. For classical argument, we let $p^k>X$; thus, we have $\mathbf{x}=\mathbf{y}$. It follows that the mean value we started with is \[ \ll k!\,p^{\frac12 k(k-1)} X^k J_{s,k}\!\left(\frac{X}{p^c}\right). \]
Thus, by choosing an appropriate $p$ and also noticing that such a mean value is closely related to $J_{s+k}(X)$, we started to build an iterative step. And if we replace $f(\boldsymbol{\alpha};X)$ with $f_k(\boldsymbol{\alpha};\boldsymbol{\eta})$, it provides a better estimation as we shall consider $\mathbf{x}\equiv \mathbf{y} \pmod{p^{k^2}}$ with multiplicity $k! \, p^{\frac12 (k+1)k(k-1)}$, and $J_s(X/p^k)$ in return by combining it with Hölder's inequality.
Eventually, when we work on a mean value, we could repeatedly multiply a $k$ factor by its congruence condition to create a new mean value. After all, if we can choose the initial condition and the recurrence condition nicely enough, we have an iterative process to bound.
Another way to utilize the efficient congruence condition is to explicitly ``nested'' within a mean value. We let $\varphi_j\in\Z[t]$ be $p^c$–spaced, meaning that \[ \varphi_j(t) \equiv t^j \pmod{p^c} \qquad (1\le j\le k). \] This family of systems $\varphi$ satisfies a translation–dilation property. While we are dealing with the number of integral solutions of \[ \sum_{i=1}^k \varphi_j\bigl(p^h z_i+\xi\bigr) \equiv \sum_{i=1}^k \varphi_j\bigl(p^h y_i+\xi\bigr) \pmod{p^B} \qquad (1\le j\le k). \]
Then, for suitable $\psi_j\in \Z[t]$, one may write \[ \varphi_j(t) = \sum_{i=1}^k \omega_{ij} t^i + p^c t^{k+1} \psi_j(t) \qquad (1\le j\le k) \] for some integral coefficient matrix $A = (\omega_{ij}) \equiv I_k \pmod{p^c}$. Then $\det(A)\not\equiv 0 \pmod{p^c}$, $\det(A)\not\equiv 0 \pmod{p^B}$ for $B>c$, i.e.\ $A^{-1}$ modulo $p^B$ exists in $\Z/p^B\Z$. So that replacing $\varphi_j$ and $\psi_j$ with $A^{-1}\varphi_j$ and $A^{-1}\psi_j$, respectively, we deduce that \[ \varphi_j(t) = t^j + p^c t^{k+1} \psi_j(t) \qquad (1\le j\le k). \]
Thus, by replacing $t$ with $p^h t+\xi$ and taking a binomial expansion, we find $\tilde{\gamma}_j\in\Z[t]$ and \[ \varphi_j(p^h t+\xi) - \varphi_j(\xi) = \Phi_j(p^h t), \] in which \[ \Phi_j(t) = \sum_{i=1}^k \Omega_{ij} t^i + p^c t^{k+1} \Psi_j(t) \qquad (1\le j\le k), \] and the integral coefficients satisfy \[ \Omega_{ij} \equiv \binom{j}{i} \pmod{p^c} \qquad (1\le i,j\le k). \] Here, we adopt the convention that $\binom{i}{j} = 0$ if $i>j$.
Similarly, there would be $\Upsilon_j\in\Z[t]$ satisfying the initial congruence condition; then \[ \sum_{i=1}^k (p^h)^j \bigl( \Phi_j(y_i) - \Phi_j(z_i) \bigr) \equiv 0 \pmod{p^B} \qquad (1\le j\le k), \] in which $\Phi_j$ has been refined by \[ \Phi_j(t) = t^j + p^{c+h} t^{k+1} \Upsilon_j(t). \]
It follows that the congruence is \[ \sum_{i=1}^k \Phi_j(y_i) \equiv \sum_{i=1}^k \Phi_j(z_i) \pmod{p^{B-hk}} \qquad (1\le j\le k). \] Therefore, whenever the original congruence condition is valid, this congruence condition follows implicitly. If we consider this as a mean value explicitly within the mean value that counts the solutions of the original congruence, it does not affect the value.
Therefore, if we sum over $\boldsymbol{\eta}$ with $\mathbf{y}\equiv\boldsymbol{\eta}\pmod{p^{H-h}}$, we have, for any $1\le\xi\le p^H$, $\xi = p^h \eta + \zeta$, indicating the summation over $p^h\eta + \xi \equiv \zeta \pmod{p^H}$ for all $y\in\Z$. Nevertheless, for the mean value subject to the implied congruence, it is considered to be satisfied; thus, letting $\boldsymbol{\beta} = 0$ for $e(\beta_1\Phi_1(\eta_i)+\cdots+\beta_k\Phi_k(\eta_i))$ is not problematic. Thus, we transfer the congruence modulo $p^h$ to modulo $p^H$.
This creates an iteration path. Moreover, if we take the same approach for any rational function whose Wronskian is non–zero modulo $p$, and we sum over an arithmetic progression modulo $p^c$, then we can link this system of rational functions to the system of $p^c$–spaced functions.
Hence, it completes the proof.