[proofplan] The proof is the classical Hardy-Littlewood circle method for Waring's problem in sufficiently many variables. First one rewrites the representation function as a Fourier coefficient of the $k$-th power Weyl sum, then decomposes the unit circle into major and minor arcs. The minor arcs are negligible once $s$ is large enough, while the major arcs are evaluated by replacing the Weyl sum with a complete exponential sum times a singular integral. Absolute convergence permits the singular series to be rearranged into its Euler product of $p$-adic local densities, and the stated lower bound follows by inserting the assumed positive lower bound for that product into the asymptotic formula. [/proofplan]

[step:Express the representation count as a circle integral]Let $\mathcal L^1$ denote the one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R$. For $P\in\mathbb N$, define the Weyl sum as the map $f_k(\cdot;P):\mathbb R\to \mathbb C$ by \begin{align*} f_k(\alpha;P):=\sum_{1\le x\le P}e(\alpha x^k). \end{align*} For each $n\in\mathbb N$, put \begin{align*} P_n:=\lfloor n^{1/k}\rfloor. \end{align*} If $x_1^k+\cdots+x_s^k=n$ with $x_i\in\mathbb N$, then $1\le x_i\le P_n$ for every $1\le i\le s$. Expanding the finite product under the integral and interchanging the finite sum with the integral gives \begin{align*} \int_0^1 f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P_n\}^s}\int_0^1 e\left((x_1^k+\cdots+x_s^k-n)\alpha\right)\,d\mathcal L^1(\alpha). \end{align*} For $m\in\mathbb Z$, define $I_m:=\int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha)$. If $m=0$, then $I_m=1$. If $m\ne 0$, then direct integration gives $I_m=(e(m)-1)/(2\pi i m)=0$, since $m\in\mathbb Z$ implies $e(m)=1$. Therefore the integral counts exactly those tuples for which $x_1^k+\cdots+x_s^k=n$, and hence \begin{align*} R_{s,k}(n)=\int_0^1 f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha). \end{align*}[/step]

[guided]We first remove the arithmetic equation by Fourier orthogonality. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure. For $P\in\mathbb N$, define the map $f_k(\cdot;P):\mathbb R\to\mathbb C$ by \begin{align*} f_k(\alpha;P):=\sum_{1\le x\le P}e(\alpha x^k). \end{align*} For the fixed integer $n$, set \begin{align*} P_n:=\lfloor n^{1/k}\rfloor. \end{align*} Every positive integer solution of \begin{align*} x_1^k+\cdots+x_s^k=n \end{align*} automatically satisfies $x_i^k\le n$, hence $1\le x_i\le P_n$. Thus counting solutions in $\mathbb N^s$ is the same as counting tuples in $\{1,\dots,P_n\}^s$ satisfying the same equation. We now prove the orthogonality identity directly. Since the sum defining $f_k(\alpha;P_n)$ is finite, expanding the $s$-th power and interchanging the resulting finite sum with the integral is justified without any convergence theorem. We obtain \begin{align*} \int_0^1 f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=\sum_{(x_1,\dots,x_s)\in\{1,\dots,P_n\}^s}\int_0^1 e\left((x_1^k+\cdots+x_s^k-n)\alpha\right)\,d\mathcal L^1(\alpha). \end{align*} For an integer $m$, define \begin{align*} I_m:=\int_0^1 e(m\alpha)\,d\mathcal L^1(\alpha). \end{align*} If $m=0$, then $I_m=1$. If $m\ne 0$, then $e(m\alpha)=\exp(2\pi i m\alpha)$ and direct integration gives \begin{align*} I_m=\frac{\exp(2\pi i m)-1}{2\pi i m}=0, \end{align*} because $m\in\mathbb Z$ implies $\exp(2\pi i m)=1$. Thus the inner integral is $1$ exactly when $x_1^k+\cdots+x_s^k=n$ and is $0$ otherwise. Since every such solution has $1\le x_i\le P_n$, the sum of these indicators is precisely $R_{s,k}(n)$. Hence \begin{align*} R_{s,k}(n)=\int_0^1 f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha). \end{align*} This identity is the entry point for the circle method: the arithmetic condition has become the extraction of the $n$-th Fourier coefficient of the generating exponential sum.[/guided]

[step:Choose enough variables so the minor arcs are negligible]By the defining property of the chosen threshold $s_0(k)$, for every $s\ge s_0(k)$ the unit interval has a measurable major-arc set $\mathfrak M_n\subset[0,1]$ and complementary minor-arc set $\mathfrak m_n:=[0,1]\setminus\mathfrak M_n$ such that \begin{align*} \int_{\mathfrak m_n} f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=o(P_n^{s-k}) \end{align*} as $n\to\infty$. This is exactly the minor-arc hypothesis included in the statement, with $P=P_n$.[/step]

[guided]The minor arcs are not being estimated from scratch in this theorem; their negligible contribution is part of the admissibility property required of $s_0(k)$. Since $s\ge s_0(k)$ and $P_n=\lfloor n^{1/k}\rfloor$, that hypothesis provides a measurable set $\mathfrak M_n\subset[0,1]$ and its complement $\mathfrak m_n=[0,1]\setminus\mathfrak M_n$. Applying the stated minor-arc estimate with this cutoff gives \begin{align*} \int_{\mathfrak m_n} f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha)=o(P_n^{s-k}). \end{align*} This is the precise point where the condition that there are sufficiently many variables is used: it makes the contribution away from rational approximations small compared with the expected main scale.[/guided]

[step:Evaluate the major arcs by complete sums and the singular integral]For $P>0$, define the oscillatory integral as the map $v_k(\cdot;P):\mathbb R\to \mathbb C$ by \begin{align*} v_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t). \end{align*} For $P>0$ and $n\in\mathbb N$, define the singular integral factor as the map $\mathfrak J_{s,k}:\mathbb N\times(0,\infty)\to \mathbb C$ by \begin{align*} \mathfrak J_{s,k}(n;P):=\int_{-\infty}^{\infty}v_k(\beta;P)^s e(-n\beta)\,d\mathcal L^1(\beta). \end{align*} By the integrated major-arc hypothesis in the definition of $s_0(k)$, applied with $P=P_n$, the major arcs satisfy \begin{align*} \int_{\mathfrak M_n} f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha) =\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)+o(P_n^{s-k}). \end{align*}[/step]

[guided]The statement has packaged the major-arc analysis as a precise hypothesis on the threshold $s_0(k)$. We first define the continuous factor that appears in that hypothesis. For $P>0$, let $v_k(\cdot;P):\mathbb R\to\mathbb C$ be the map \begin{align*} v_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t). \end{align*} For $P>0$ and $n\in\mathbb N$, define \begin{align*} \mathfrak J_{s,k}(n;P):=\int_{-\infty}^{\infty}v_k(\beta;P)^s e(-n\beta)\,d\mathcal L^1(\beta). \end{align*} Since $s\ge s_0(k)$, the integrated major-arc hypothesis applies to the major-arc set $\mathfrak M_n$ and the cutoff $P_n=\lfloor n^{1/k}\rfloor$. It gives directly \begin{align*} \int_{\mathfrak M_n} f_k(\alpha;P_n)^s e(-n\alpha)\,d\mathcal L^1(\alpha) =\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)+o(P_n^{s-k}). \end{align*} This is the point at which the arithmetic complete sums, the disjoint major arcs, and the truncation estimates have already been accounted for by the stated hypothesis; the present proof only uses the resulting integrated estimate.[/guided]

[step:Evaluate the singular integral and replace the cutoff by $n^{1/k}$]The singular integral evaluation is one of the hypotheses in the definition of $s_0(k)$. Applying it with $P=P_n$ gives \begin{align*} \mathfrak J_{s,k}(n;P_n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}n^{s/k-1}+o(n^{s/k-1}). \end{align*} Since $P_n=\lfloor n^{1/k}\rfloor$, this is the real-density contribution associated with positive $k$-th powers and the cutoff used in the circle integral.[/step]

[guided]The admissibility hypothesis for $s_0(k)$ also includes the real-density computation. We apply it with the same cutoff $P_n=\lfloor n^{1/k}\rfloor$ used in the Fourier coefficient identity and in the major-arc approximation. It yields \begin{align*} \mathfrak J_{s,k}(n;P_n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}n^{s/k-1}+o(n^{s/k-1}). \end{align*} The factor involving the Gamma function is the singular integral contribution for positive $k$-th powers; the remaining dependence on $n$ has the scaling $n^{s/k-1}$ expected from the dilation $x_i=n^{1/k}y_i$.[/guided]

[step:Combine the major and minor arc estimates]Let $s_0(k)$ be the threshold specified in the theorem statement. Combining the circle integral identity, the decomposition $[0,1]=\mathfrak M_n\cup\mathfrak m_n$, the minor-arc estimate, and the major-arc evaluation gives \begin{align*} R_{s,k}(n)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)+o(P_n^{s-k}). \end{align*} By the singular-integral evaluation and the uniform boundedness of $\mathfrak S_{s,k}(n)$ for fixed $s$ and $k$, \begin{align*} \mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}\mathfrak S_{s,k}(n)n^{s/k-1}+o(n^{s/k-1}). \end{align*} Since $P_n=\lfloor n^{1/k}\rfloor$, one has $P_n\le n^{1/k}$ and $P_n\ge 2^{-1}n^{1/k}$ for all sufficiently large $n$, so $o(P_n^{s-k})=o(n^{s/k-1})$. Therefore \begin{align*} R_{s,k}(n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}\mathfrak S_{s,k}(n)n^{s/k-1}+o(n^{s/k-1}). \end{align*} This proves the asserted Hardy-Littlewood asymptotic formula.[/step]

[guided]We now combine the two pieces of the circle integral. The identity from the first step and the disjoint decomposition $[0,1]=\mathfrak M_n\cup\mathfrak m_n$ give the sum of the major-arc and minor-arc integrals. Substituting the two estimates already obtained gives \begin{align*} R_{s,k}(n)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)+o(P_n^{s-k}). \end{align*} Next insert the singular-integral evaluation: \begin{align*} \mathfrak J_{s,k}(n;P_n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}n^{s/k-1}+o(n^{s/k-1}). \end{align*} The theorem statement includes uniform boundedness of $\mathfrak S_{s,k}(n)$ for fixed $s$ and $k$, so multiplying the $o(n^{s/k-1})$ error by $\mathfrak S_{s,k}(n)$ still gives $o(n^{s/k-1})$. Hence \begin{align*} \mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P_n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}\mathfrak S_{s,k}(n)n^{s/k-1}+o(n^{s/k-1}). \end{align*} It remains only to compare the minor-arc scale with the displayed main scale. Since $P_n=\lfloor n^{1/k}\rfloor$, for all sufficiently large $n$ we have $2^{-1}n^{1/k}\le P_n\le n^{1/k}$. Therefore $P_n^{s-k}$ is comparable to $n^{s/k-1}$, and $o(P_n^{s-k})=o(n^{s/k-1})$. Combining these estimates gives \begin{align*} R_{s,k}(n)=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}\mathfrak S_{s,k}(n)n^{s/k-1}+o(n^{s/k-1}). \end{align*}[/guided]

[step:Identify the singular series with the product of local densities]For $q\in\mathbb N$, define the congruence-counting function as the map $M_{s,k}:\mathbb N\times\mathbb N\to\mathbb N$ by \begin{align*} M_{s,k}(n,q):=\#\left\{x\in(\mathbb Z/q\mathbb Z)^s:x_1^k+\cdots+x_s^k\equiv n\pmod q\right\}. \end{align*} By the local-density Euler-product hypothesis in the definition of $s_0(k)$, each limit \begin{align*} \sigma_p(n)=\lim_{h\to\infty}p^{h(1-s)}M_{s,k}(n,p^h) \end{align*} exists, the Euler product converges, and \begin{align*} \mathfrak S_{s,k}(n)=\prod_p \sigma_p(n). \end{align*}[/step]

[guided]The congruence counts used in the local factors must match the statement. For $q\in\mathbb N$, define $M_{s,k}:\mathbb N\times\mathbb N\to\mathbb N$ by \begin{align*} M_{s,k}(n,q):=\#\left\{x\in(\mathbb Z/q\mathbb Z)^s:x_1^k+\cdots+x_s^k\equiv n\pmod q\right\}. \end{align*} The admissibility property of $s_0(k)$ includes the local-density Euler product for the absolutely convergent singular series. Applying that hypothesis gives, for every prime $p$ and every $n\in\mathbb N$, the existence of \begin{align*} \sigma_p(n)=\lim_{h\to\infty}p^{h(1-s)}M_{s,k}(n,p^h). \end{align*} It also gives convergence of the product over primes and the identity \begin{align*} \mathfrak S_{s,k}(n)=\prod_p \sigma_p(n). \end{align*}[/guided]

[step:Deduce the uniform positive lower bound on representations]Fix $s\ge s_0(k)$, a set $E\subset\mathbb N$, and a constant $c_{s,k}>0$ such that \begin{align*} \prod_p\sigma_p(n)\ge c_{s,k} \end{align*} for every $n\in E$. Define \begin{align*} A_{s,k}:=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}. \end{align*} Then $A_{s,k}>0$, and the asymptotic formula together with the Euler-product identity gives, uniformly as $n\to\infty$ through positive integers, \begin{align*} R_{s,k}(n)=A_{s,k}\left(\prod_p\sigma_p(n)\right)n^{s/k-1}+o(n^{s/k-1}). \end{align*} Hence there is $N_{s,k}\in\mathbb N$ such that, for every $n\ge N_{s,k}$, \begin{align*} |o(n^{s/k-1})|\le \frac{1}{2}A_{s,k}c_{s,k}n^{s/k-1}. \end{align*} For every $n\in E$ with $n\ge N_{s,k}$, we therefore obtain \begin{align*} R_{s,k}(n)\ge \frac{1}{2}A_{s,k}c_{s,k}n^{s/k-1}. \end{align*} Thus the conclusion holds with \begin{align*} C_{s,k}:=\frac{1}{2}\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}c_{s,k}. \end{align*} This completes the proof.[/step]

[guided]Fix $s\ge s_0(k)$, a set $E\subset\mathbb N$, and $c_{s,k}>0$ such that \begin{align*} \prod_p\sigma_p(n)\ge c_{s,k} \end{align*} for every $n\in E$. Define the positive real constant \begin{align*} A_{s,k}:=\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}. \end{align*} The Gamma function is positive on positive real arguments, so $A_{s,k}>0$. The asymptotic formula and the identity $\mathfrak S_{s,k}(n)=\prod_p\sigma_p(n)$ give \begin{align*} R_{s,k}(n)=A_{s,k}\left(\prod_p\sigma_p(n)\right)n^{s/k-1}+o(n^{s/k-1}). \end{align*} By the definition of the little-$o$ term, there exists $N_{s,k}\in\mathbb N$ such that for every $n\ge N_{s,k}$ its absolute value is at most \begin{align*} \frac{1}{2}A_{s,k}c_{s,k}n^{s/k-1}. \end{align*} If $n\in E$ and $n\ge N_{s,k}$, then the assumed lower bound for the local-density product gives \begin{align*} R_{s,k}(n)\ge A_{s,k}c_{s,k}n^{s/k-1}-\frac{1}{2}A_{s,k}c_{s,k}n^{s/k-1}. \end{align*} Thus \begin{align*} R_{s,k}(n)\ge \frac{1}{2}A_{s,k}c_{s,k}n^{s/k-1}. \end{align*} The required constant may therefore be chosen as \begin{align*} C_{s,k}:=\frac{1}{2}\frac{\Gamma(1+1/k)^s}{\Gamma(s/k)}c_{s,k}. \end{align*}[/guided]