10. Lecture 10  PDF

Vinogradov’s system of diophantine equations

• Let \(k, n \geqslant 1\) be integers, and \(P\in\mathbb Z_+\) be a large integer. Let \(J_{k, n}(P)\) be the number of integer solutions of the system of diophantine equations \[\begin{align*} & x_{1}+\cdots+x_{k}-x_{k+1}-\cdots-x_{2 k}=0 \\ & x_{1}^2+\cdots+x_{k}^2-x_{k+1}^2-\cdots-x_{2 k}^2=0 \\ & \hspace{2.05cm}\vdots \hspace{2.65cm}\vdots \tag{HE}\\ & x_{1}^{n}+\cdots+x_{k}^{n}-x_{k+1}^{n}-\cdots-x_{2 k}^{n}= 0, \end{align*}\] where \(1 \leqslant x_{1}, \ldots, x_{2 k} \leqslant P\).

• More generally, for given integers \(\lambda_{1}, \ldots, \lambda_{n}\in\mathbb Z\), let us define \(J_{k, n}\left(P; \lambda_{1}, \ldots, \lambda_{n}\right)\) as the number of solutions of the system \[\begin{align*} & x_{1}+\cdots+x_{k}-x_{k+1}-\cdots-x_{2 k}=\lambda_1 \\ & x_{1}^2+\cdots+x_{k}^2-x_{k+1}^2-\cdots-x_{2 k}^2=\lambda_2 \\ & \hspace{2.05cm}\vdots \hspace{2.65cm}\vdots \tag{IE}\\ & x_{1}^{n}+\cdots+x_{k}^{n}-x_{k+1}^{n}-\cdots-x_{2 k}^{n}= \lambda_n, \end{align*}\] where \(1 \leqslant x_{1}, \ldots, x_{2 k} \leqslant P\). Then \(J_{k, n}(P)=J_{k, n}\left(P; 0, \ldots, 0\right)\).

Integral representation of \(J_{k, n}\left(P; \lambda_{1}, \ldots, \lambda_{n}\right)\)

• The basis of this method is the elementary orthogonality identity \[\begin{aligned} \int_{0}^{1} e(k x) d x=\int_{0}^{1} e^{2 \pi i k x} d x= \begin{cases} 1 & \text { if } k=0,\\ 0 & \text { if } k\neq0. \end{cases} \end{aligned}\]

• Using this identity we note that \[\begin{gathered} J_{k, n}\left(P; \lambda_{1}, \ldots, \lambda_{n}\right)=\sum_{1 \leqslant x_{1}, \ldots, x_{2 k} \leqslant P} \prod_{m=1}^n\delta_{\lambda_m}\Big(\sum_{j=1}^kx_j^m-\sum_{j=k+1}^{2k}x_j^m\Big)\\ =\sum_{1 \leqslant x_{1}, \ldots, x_{2 k} \leqslant P} \prod_{m=1}^n\int_0^1e\Big(\Big(\sum_{j=1}^kx_j^m-\sum_{j=k+1}^{2k}x_j^m-\lambda_m\Big)\alpha_m\Big)d\alpha_m\\ \int_{0}^{1} \cdots \int_{0}^{1}\Big|\sum_{x \in [P]} e\left(\alpha_{1} x+\cdots+\alpha_{n} x^{n}\right)\Big|^{2 k} e\left(-\alpha_{1} \lambda_{1}-\cdots-\alpha_{n} \lambda_{n}\right) d \alpha_{1} \cdots d \alpha_{n}. \end{gathered}\]

• We then immediately see that \[J_{k, n}\left(P; \lambda_{1}, \ldots, \lambda_{n}\right)\le J_{k, n}(P),\] by taking \(\lambda_1=\ldots=\lambda_n=0\).

Simple properties of \(J_{k, n}\left(P; \lambda_{1}, \ldots, \lambda_{n}\right)\)

• When \(x_{1}, \ldots, x_{2 k}\) run over all possible \(P^{2 k}\) values, then the left-hand side of the system (IE) assumes all possible values \(\lambda_{1}, \ldots, \lambda_{n}\), which satisfy \[\left|\lambda_{1}\right|<k P,\left|\lambda_{2}\right|<k P^{2}, \ldots,\left|\lambda_{n}\right|<k P^{n}.\]

• By the Fourier inverse transform we have \[\begin{aligned} & \Big|\sum_{x \leqslant P} e\left(\alpha_{1} x+\cdots+\alpha_{n} x^{n}\right)\Big|^{2 k} \\ & \quad=\sum_{\left|\lambda_{1}\right|<k P, \ldots,\left|\lambda_{n}\right|<k P^{n}} J_{k, n}\left(\lambda_{1}, \ldots, \lambda_{n}\right) e\left(-\alpha_{1} \lambda_{1}-\cdots-\alpha_{n} \lambda_{n}\right). \end{aligned}\]

• Now taking \(\alpha_1=\ldots=\alpha_n=0\) in the above equation we obtain \[\sum_{\left|\lambda_{1}\right|<k P, \ldots,\left|\lambda_{n}\right|<k P^{n}} J_{k, n}\left(\lambda_{1}, \ldots, \lambda_{n}\right)=P^{2 k}.\]

• Further, we have trivially \(J_{k, n}(P) \leqslant P^{2 k}\), and moreover \(J_{k, n}(P)\) is clearly nondecreasing as a function of \(k\) or \(P\).

• Our interest will be primarily in the upper bounds for \(J_{k, n}(P)\), but we may note here that a lower bound may be obtained as follows. \[\begin{aligned} P^{2 k} & =\sum_{\left|\lambda_{1}\right|<k P, \ldots,\left|\lambda_{n}\right|<k P^{n}} J_{k, n}\left(\lambda_{1}, \ldots, \lambda_{n}\right) \\ & \leqslant J_{k, n}(P) \sum_{\left|\lambda_{1}\right|<k P, \ldots,\left|\lambda_{n}\right|<k P^{n}} 1 \leqslant J_{k, n}(P)(2 k) P \cdots(2 k) P^{n} \\ & =J_{k, n}(P)(2 k)^{n} P^{n(n+1) / 2}, \end{aligned}\] which gives \[J_{k, n}(P) \geqslant(2 k)^{-n} P^{2 k-n(n+1) / 2},\] and this is a nontrivial bound if \(k>\frac{1}{4}\left(n^{2}+n\right)\).

• If we consider the diagonal solutions \(x_j=x_{j+k}\) for all \(j\in[k]\), and \(1\le x_1,\ldots, x_k\le P\), then \(J_{k, n}(P)\ge P^n\).

• If we consider only the first \(n-1\) equations in (IE), then the number of their solutions is \(J_{k, n-1}\left(\lambda_{1}, \ldots, \lambda_{n-1}\right)\), and if we let \(\left|\lambda_{n}\right|\) take all possible values ( \(<k P^{n}\) ) in the last equation in (IE), then we obtain \[\sum_{\left|\lambda_{n}\right|<k P^{n}} J_{k, n}\left(\lambda_{1}, \ldots, \lambda_{n}\right)=J_{k, n-1}\left(\lambda_{1}, \ldots, \lambda_{n-1}\right).\]

Linnik’s lemma

• We can also assume that \(m=1\). If \((x_{1}, \ldots, x_{n})\) is a solution of (LE) such that \(0\le x_j< p^n\) for \(j\in [n]\), then \((x_1+l_1p^n, \ldots, x_n+l_np^n)\) with \(l_1,\ldots, l_n\in[m]\) is also a solution of (LE), and there are \(m^n\) such solutions.

• For each \(\lambda\in\mathbb Z\) and \(j\in[n]\) there is \(p^{n-j}\) choices of the residue class \(\mu\pmod{p^n}\) such that \(\lambda\equiv \mu\pmod{p^j}\).

• Thus for any given tuple of integers \(\left(\lambda_{1}, \ldots, \lambda_{n}\right)\in\mathbb Z^n\), there are \(\prod_{j=1}^{n-1} p^{n-j}=p^{n(n-1) / 2}\) different vectors \(\left(\mu_{1}, \ldots, \mu_{n}\right)\in \mathbb Z/p^{n}\mathbb Z\) so that \[\mu_{j} \equiv \lambda_{j} \quad \pmod {p^{j}} \quad \text{ for all } \quad j\in[n].\]

• It will suffices to prove that for any fixed vector \(\left(\mu_{1}, \ldots, \mu_{n}\right)\in \mathbb Z/p^{n}\mathbb Z\) there are at most \(n!\) solutions \((x_1,\ldots, x_n)\in \mathbb Z/p^{n}\mathbb Z\) that the \(x_j\) are distinct \(\pmod p\) and satisfy \[\begin{align*} x_{1}+\cdots+x_{n} &\equiv \mu_{1} \pmod {p^n}, \\ x_{1}^{2}+\cdots+x_{n}^{2} &\equiv \mu_{2}\pmod {p^{n}}, \\ &\hspace{0.2cm} \vdots\tag{LEM}\\ x_{1}^{n}+\cdots+x_{n}^{n} &\equiv \mu_{n}\pmod {p^{n}}. \end{align*}\] We have “lifted” all of our congruences to be congruences modulo \(p^{n}\).

• Recall the Girard–Newton formulae. For \(k \in\mathbb Z_+\), let \[p_k(x_1, \ldots, x_n) = \sum_{i=1}^{n} x_i^k = x_1^k + \cdots + x_n^k.\]

• For \(k \in\mathbb N\), let \(e_k(x_1, \ldots, x_n)\) be the elementary symmetric polynomial \[\begin{aligned} e_0(x_1, \ldots, x_n) & = 1, \\ e_1(x_1, \ldots, x_n) & = x_1 + x_2 + \cdots + x_n, \\ e_2(x_1, \ldots, x_n) & = \sum_{1 \leq i < j \leq n} x_i x_j, \\ & \hspace{0.2cm} \vdots \\ e_n(x_1, \ldots, x_n) & = x_1 x_2 \cdots x_n, \\ e_k(x_1, \ldots, x_n) & = 0, \quad \text{for } k > n. \end{aligned}\]

• Then Newton’s identities can be stated as \[k e_k(x_1, \ldots, x_n) = \sum_{i=1}^{k} (-1)^{i-1} e_{k-i}(x_1, \ldots, x_n) p_i(x_1, \ldots, x_n),\] valid for all \(n \geq k \geq 1\).

• From (LEM) we see that \(p_i(x_1, \ldots, x_n)\equiv \mu_i\pmod {p^n}\) for \(i\in[n]\).

• Since \((p, n!)=1\), then the elementary functions \(e_j=e_j(x_1, \ldots, x_n)\) given as solutions \(\pmod {p^n}\) of the following linear equations \[\begin{gathered} e_1=p_1, 2e_2=(e_1p_1-p_2), 3e_3=(e_2p_1-e_1p_2+p_3), \\ 4e_4=(e_3p_1-e_2p_2+e_1p_3-p_4), \ldots, n e_n = \sum_{i=1}^{n} (-1)^{i-1} e_{n-i} p_i, \end{gathered}\] are uniquely determined by \(\mu_i\pmod {p^n}\) for \(i\in[n]\).

• We also know that the polynomial with roots \(x_1,\ldots, x_n\) may be expressed as \[\begin{aligned} \prod_{i\in[n]}(x-x_i)=\sum_{k=0}^n(-1)^ke_k(x_1,\ldots, x_n)x^{n-k}. \end{aligned}\]

• Therefore, the polynomial \(\prod_{i\in[n]}(x-x_i)\) in also uniquely determined by \(\mu_i\pmod {p^n}\) for \(i\in[n]\).

• Now suppose that there are two solutions \((x_1,\ldots, x_n)\in \mathbb Z/p^{n}\mathbb Z\) and \((y_1,\ldots, y_n)\in \mathbb Z/p^{n}\mathbb Z\) with distinct entries \(\pmod p\) such that \[\begin{aligned} \sum_{j\in[n]}x_j^k\equiv \sum_{j\in[n]}y_j^k\equiv\mu_k\pmod{p^n} \quad \text{ for all }\quad k\in[n], \end{aligned}\] then we show that \((y_1,\ldots, y_n)\) is a permutation of \((x_1,\ldots, x_n)\).

• By the previous discussion the polynomials \[\begin{aligned} P(z)=\prod_{i\in[n]}(z-x_i) \quad \text{ and } \quad Q(z)=\prod_{i\in[n]}(z-y_i) \end{aligned}\] are identically congruent \(\pmod{p^n}\).

• But we have \(P\left(x_{j}\right) \equiv 0\pmod {p^{n}}\) for all \(j\in[n]\), and so we must have \[Q\left(x_{j}\right)=\prod_{i=1}^{n}\left(x_{j}-y_{i}\right) \equiv 0 \pmod {p^{n}} \quad \text{ for all } \quad j\in[n].\]

• If the \(y_{i}\) are distinct modulo \(p\) this implies that \(x_{j}\) is congruent to one of the \(y_{i}\pmod {p^{n}}\), and so (since the \(x_{j}\) are also distinct modulo \(p\)) the \(x_{j}\) are forced to be a permutation of the \(y_{j}\pmod p{^{n}}\). This implies that there are at most \(n!\) possible solution vectors \((x_1,\ldots, x_n)\). $$\tag*{$\blacksquare$}$$

Recursive estimate

We now formulate a recursive estimate for \(J_{k, n}(P)\), which will enable us to bound it explicitly. This is the crucial part of the Vinogradov–Korobov method.

• Let us also note that \(p>n\), because of our hypothesis that \(P \geq(2 n)^{3 n}\). We will be able to apply Linnik’s lemma to \(n\) of our variables.

• We choose \(p \simeq P^{1 / n}\) so that the ranges \(m p^{n}\) of the variables in Linnik’s lemma will approximately match the ranges \(P\) of our variables.

• Next, let \(J_{1}\) denote the number of solution vectors \(\left(x_{1}, \ldots, x_{2 k}\right)\), counted by \(J_{k, n}\left(p P_{1}\right)\), in which \(\left(x_{1}, \ldots, x_{k}\right)\) and \(\left(x_{k+1}, \ldots, x_{2 k}\right)\) each contain \(n\) numbers that are distinct modulo \(p\), and let \(J_{2}\) denote the number of solution vectors not counted by \(J_{1}\).

• Then it suffices to estimate \(J_1\) and \(J_2\) separately, since \[J_{k, n}\left(p P_{1}\right)=J_{1}+J_{2}.\]

• Also let \(J_{1}^{\prime}\) denote the number of solution vectors \(\left(x_{1}, \ldots, x_{2 k}\right)\), counted by \(J_{k, n}\left(p P_{1}\right)\), for which the first \(n\) elements \(\left(x_{1}, \ldots, x_{n}\right)\) and \(\left(x_{k+1}, \ldots, x_{k+n}\right)\) are distinct modulo \(p\). Then we have \[J_{k, n}\left(p P_{1}\right)=J_{1}+J_{2} \leq k^{2 n} J_{1}^{\prime}+J_{2},\] since each vector counted by \(J_{1}^{\prime}\) corresponds to at most \(k^{2 n}\) vectors counted by \(J_{1}\). This is by noting that the first \(n\) entries of a vector from \(J_1'\) may be placed in \(k(k - 1) \cdots (k - n + 1)\) ways in \(k\) places without changing the order of the remaining \(k-n\) entries of the vector.

• We now prove that \[J_1'\le \max_{x\in[p]}J_1'(x),\] where \(J_{1}^{\prime}(x)\) denotes the number of solution vectors \(\left(x_{1}, \ldots, x_{2 k}\right)\), counted by \(J_{1}^{\prime}\), for which all of the \(2 k-2 n\) components \(\left(x_{n+1}, \ldots, x_{k}\right)\) and \(\left(x_{k+n+1}, \ldots, x_{2 k}\right)\) are congruent to \(x\pmod p\).

• Indeed, for \(\alpha=(\alpha_1,\ldots, \alpha_n)\in\mathbb [0, 1)^n\), let \[S_{x}(\alpha)=\sum_{\substack{{z=1}\\z\equiv x(\bmod p)}}^{P_1}e(\alpha_1z+\ldots+\alpha_nz^n),\] and observe that \[\begin{aligned} J_{1}^{\prime}=\int_{[0, 1)^n}\Big|\sum_{\substack{x_{1}, \ldots, x_{n}\in[p]\\{\rm distinct}}} S_{x_1}\left(\alpha\right) \cdots S_{x_n}\left(\alpha\right)\Big|^{2}\Big|\sum_{x \in [p]} S_x(\alpha)\Big|^{2 k-2 n} d \alpha. \end{aligned}\]

• Using Hölder’s inequality, pulling out the inner sum \(\sum_{x\in[p]}\) and taking \(\max_{x\in[p]}\) we obtain \[\begin{aligned} J_{1}^{\prime}\le p^{2k-2n}\max_{x\in[p]}\int_{[0, 1)^n}\Big|\sum_{\substack{x_{1}, \ldots, x_{n}\in[p]\\{\rm distinct}}} S_{x_1}\left(\alpha\right) \cdots S_{x_n}\left(\alpha\right)\Big|^{2}|S_x(\alpha)|^{2 k-2 n} d \alpha. \end{aligned}\]

• For every \(x\in[p]\), the last integral is precisely equal to \(J_1'(x)\) as desired.

• If \((x_1,\ldots, x_{2k})\) is a solution counted by \(J_1'(x)\), then it has the form \[\begin{aligned} &(x_1,\ldots,x_n,x_{n+1},\ldots, x_{k}, x_{k+1},\ldots, x_{k+n}, x_{k+n+1},\ldots, x_{2k})\\ &=(x_1,\ldots,x_n,py_{n+1}+x,\ldots, py_{k}+x, x_{k+1},\ldots, x_{k+n},py_{k+n+1}+x,\ldots, py_{2k}+x), \end{aligned}\] where \(\left(x_{1}, \ldots, x_{n}\right)\) and \(\left(x_{k+1}, \ldots, x_{k+n}\right)\) are distinct modulo \(p\), and \(y_{n+1},y_{k+n+1}, \ldots, y_{k}, y_{2k}\in[P_1]\).

• Observe that the system (HE) is translation invariant, which means that if \((x_1,\ldots, x_{2k})\) satisfies (HE), then \((x_1-x,\ldots, x_{2k}-x)\) also does.

• Hence, by the translation invariance, we have \[\begin{aligned} &\sum_{i=1}^nx_{i}^j-x_{k+i}^j + \sum_{i=n+1}^{k}(py_{i}+x)^j-(py_{k+i}+x)^j=0 \quad \text{ for all }\quad j\in[n],\\ \iff& \sum_{i=1}^nz_{i}^j-z_{k+i}^j + p^j\sum_{i=n+1}^{k}y_{i}^j-y_{k+i}^j=0 \quad \text{ for all }\quad j\in[n], \end{aligned}\] where \(z_{i}=x_i-x\) and \(z_{k+i}=x_{k+i}-x\) for all \(i\in[n]\). Moreover, we have \(1-x\le z_{i}, z_{k+i}\le pP_1-x\) for \(i\in[n]\), and \(1\le y_i, y_{k+i}\le P_1\) for \(i\in[k]\setminus[n]\), and \(\left(z_{1}, \ldots, z_{n}\right)\) and \(\left(z_{k+1}, \ldots, z_{k+n}\right)\) are distinct modulo \(p\).

• The last system of equations can be rewritten as \[\begin{aligned} \sum_{i=1}^nz_{i}^j=\sum_{i=1}^nz_{k+i}^j - p^j\sum_{i=n+1}^{k}y_{i}^j-y_{k+i}^j \quad \text{ for all }\quad j\in[n]. \end{aligned}\]

• Fixing \(z_{k+1}, \ldots, z_{k+n}\), each vector \(\left(z_{1}, \ldots, z_{n}\right)\) satisfies the conditions of Linnik’s lemma, with \(A=1-x\) and \(m\ge pP_1p^{-n}>Pp^{-n}\), so that we may take \(m=\lfloor Pp^{-n}\rfloor+1\). Thus by Linnik’s lemma \[T\le n!m^np^{n(n-1)/2}.\]

• For any fixed \(z_{1}, \ldots, z_{n}\) and \(z_{k+1}, \ldots, z_{k+n}\), the number of vectors \(\left(y_{n+1}, \ldots, y_{k}, y_{k+n+1},\ldots, y_{2k}\right)\) that are counted in \(J_{1}^{\prime}(x)\) is at most \(J_{k-n, n}\left(P_{1}\right)\).

• So in total, using the trivial bound \(\left(p P_{1}\right)^{n}\) for the number of choices of \(z_{k+1}, \ldots, z_{k+n}\), we may write \[\begin{aligned} J_1\le k^{2n}J_1'(x)\le k^{2 n} p^{2 k-2 n} n!m^{n} p^{n(n-1) / 2}\left(p P_{1}\right)^{n} J_{k-n, n}\left(P_{1}\right). \end{aligned}\]

• Since \(p \geqslant \frac{1}{2} P^{1 / n}\) we have \(P p^{-n} \leqslant 2^{n}\), implying \(m^{n} \leqslant 2^{n^{2}+n} \leqslant 2^{k}\).

• Using further \(p \leqslant P^{1 / n}\), and \(P_{1} \leqslant 4 P^{(n-1) / n}\), we obtain \[\begin{aligned} J_{1} &\leqslant n!2^{k} k^{2 n} P^{2 k / n+(3 n-5) / 2} 4^{n} J_{k-n, n}\left(P_{1}\right) \\ &\leqslant \frac{1}{2} 4^{2 k} P^{2 k / n+(3 n-5) / 2} J_{k-n, n}\left(P_{1}\right), \end{aligned}\] because \(k \geqslant n^{2}+n\), and \(n \geqslant 2\).

• Recall that \(J_{2}\) counts all those vectors \(\left(x_{1}, \ldots, x_{k}, x_{k+1}, \ldots, x_{2 k}\right)\), counted by \(J_{k, n}\left(p P_{1}\right)\), in which either \(\left(x_{1}, \ldots, x_{k}\right)\) or \(\left(x_{k+1}, \ldots, x_{2 k}\right)\) contains at most \(n-1\) numbers that are distinct \(\pmod p\).

• In the first case there are at most \(p^{n-1} (n-1)^{k}\) possibilities for \(\left(x_{1}\pmod p, \ldots, x_{k}\pmod p\right)\). Indeed, there are at most \(p^{n-1}\) ways of choosing \(\{u_1,\ldots, u_k\}\subseteq \mathbb Z/p\mathbb Z\), and then there are at most \((n-1)^k\) possibilities of fixing \(\left(x_{1}\pmod p, \ldots, x_{k}\pmod p\right)\) with coordinates from \(\{u_1,\ldots, u_k\}\). Hence, there are at most \(p^{n-1+k} n^{k}\) possibilities for \(\left(x_{1}\pmod p, \ldots, x_{k}\pmod p, x_{k+1}\pmod p, \ldots, x_{2 k}\pmod p\right)\).

• We proceed similarly in the second case.

• Therefore, if \(\mathcal{A}\) denote the set of all possible vectors of the form \(\left(x_{1}\pmod p, \ldots, x_{k}\pmod p, x_{k+1}\pmod p, \ldots, x_{2 k}\pmod p\right)\) that are counted in \(J_{2}\), then \(\# \mathcal{A} \leq 2 p^{n-1+k} n^{k}\).

• Observe that \[\begin{aligned} & \bigg|\sum_{(x_{1}, \ldots, x_{2 k})\in\mathcal A} S_{x_1}(\alpha) \cdots S_{x_k}(\alpha) \overline{S_{x_{k+1}}(\alpha)\cdots S_{x_{2k}}(\alpha)}\bigg| \\ & \quad \leqslant\bigg(\sum_{(x_{1}, \ldots, x_{2 k})\in\mathcal A} |S_{x_1}(\alpha)|^{2 k}\bigg)^{1 / 2 k} \cdots\bigg(\sum_{(x_{1}, \ldots, x_{2 k})\in\mathcal A} |S_{x_{2k}}(\alpha)|^{2 k}\bigg)^{1 / 2 k} \\ & \quad \leqslant \sum_{x \in [p]}|S_x(\alpha)|^{2 k} \sum_{(x_{1}, \ldots, x_{2 k})\in\mathcal A} 1 \leqslant 2p^{k+n-1} n^{k} \sum_{x \in [p]}|S_x(\alpha)|^{2 k} \\ & \quad \leqslant 2p^{k+n-1} n^{k} P_{1}^{2 n} \sum_{x \in [p]}|S_x(\alpha)|^{2(k-n)}. \end{aligned}\] since trivially \(|S_x(\alpha)| \leqslant P_{1}\). Therefore \[\begin{aligned} J_{2}= & \int_{[0, 1)^n}\sum_{(x_{1}, \ldots, x_{2 k})\in\mathcal A} S_{x_1}(\alpha) \cdots S_{x_k}(\alpha) \overline{S_{x_{k+1}}(\alpha)\cdots S_{x_{2k}}(\alpha)}d \alpha \\ \leqslant & p^{k+n-1} n^{k} P_{1}^{2 n} \int_{[0, 1)^n} \sum_{x \in [P_{1}]}|S_x(\alpha)|^{2(k-n)} d \alpha \\ = & p^{k+n-1} n^{k} P_{1}^{2 n} J_{k-n, n}\left(P_{1}\right) \leqslant \frac{1}{2} 4^{2 k} P^{2 k / n+(3 n-5) / 2} J_{k-n, n}\left(P_{1}\right). \end{aligned}\]

• This completes the proof. $$\tag*{$\blacksquare$}$$

Vinogradov’s mean value theorem

• To bound \(J_{k-n, n}\left(P_{1}\right)\) we may use the the induction hypothesis, since \[k-n \geqslant n^{2}+n m, \quad \text{ and } \quad P_{1} \geqslant P^{1-1 / n} \geqslant(2 n)^{3 n(1+1 /(n-1))^{m}} .\]

• It follows that \[\begin{aligned} J_{k-n, n}\left(P_{1}\right) & \leqslant(4 n)^{4(k-n) m} P_{1}^{2(k-n)-(n^{2}+n) / 2+c_{m}} \\ & \leqslant(4 n)^{4(k-n) m} 4^{2(k-n)} P^{(1-1 / n)(2 k-2 n-(n^{2}+n) / 2+c_{m})}, \end{aligned}\] which combined with (A) gives (*).

• Let now \[P<(2 n)^{3 n(1+1 /(n-1))^{r}},\] and use induction again, supposing \[P<(2 n)^{3 n(1+1 /(n-1))^{m+1}}, \quad \text{ and }\quad k \geqslant n^{2}+n(m+1).\]

• Then we have trivially \[J_{k, n}(P) \leqslant P^{2 n} J_{k-n, n}(P), \tag{B}\] and if \(P \geqslant(2 n)^{3 n(1+1 /(n-1))^{m}}\) then by the first part of the proof \(J_{k-n, n}(P)\) may be estimated by (*) as before.

• Otherwise, we use the induction hypothesis to estimate \(J_{k-n, n}(P)\), so that we obtain in any case \[J_{k-n, n}(P) \leqslant(4 n)^{4 k m} P^{2(k-n)-(n^{2}+n)/2+c_{m}},\] and (B) gives \[J_{k, n}(P) \leqslant(4 n)^{4 k(m+1)} P^{2 k-(n^{2}+n)/2+c_{m+1}},\] since \[P<(2 n)^{3 n(1+1 /(n-1))^{m+1}}, \quad \text{ and }\quad k \geqslant n^{2}+n(m+1),\] implies \[P^{c_{m}-c_{m+1}} \leqslant(4 n)^{4 k}.\]

• This completes the proof of the Vinogradov mean value theorem. $$\tag*{$\blacksquare$}$$