[proofplan] We prove that every sufficiently large integer is a sum of $s=2^k+1$ positive $k$-th powers; allowing zero summands then gives the asserted bound for the non-negative version of $G(k)$. The circle method writes the ordered representation count as an integral of a Weyl sum. Hua's mean value estimate controls $2^k$ factors on the minor arcs, while Weyl's inequality gives a saving on the remaining factor. On the major arcs, the standard Hardy-Littlewood analysis gives a positive main term of order $P^{s-k}$, because the real density and all $p$-adic densities are positive in Hua's range. [/proofplan]

[step:Reduce the theorem to positivity of an ordered representation count] Fix an integer $k\ge 2$ and put \begin{align*} s:=2^k+1. \end{align*} For $n\in\mathbb N$, define $R_{s,k}(n)$ to be the number of ordered tuples $(x_1,\dots,x_s)\in\mathbb N^s$ such that \begin{align*} x_1^k+\cdots+x_s^k=n. \end{align*} If $R_{s,k}(n)>0$ for all sufficiently large $n$, then every sufficiently large $n$ is a sum of $s$ positive $k$-th powers. Since every positive $k$-th power is also a non-negative $k$-th power, this implies $G(k)\le s=2^k+1$. Thus it is enough to prove that there exists $n_0=n_0(k)\in\mathbb N$ such that \begin{align*} R_{s,k}(n)>0 \end{align*} for every $n\ge n_0$. [/step]

[step:Write the representation count as a circle method integral]Let $n\in\mathbb N$ and define \begin{align*} P:=\lfloor n^{1/k}\rfloor. \end{align*} Let $\mathcal L^1$ denote [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R$. Define the Weyl sum $f(\cdot;P):\mathbb R\to\mathbb C$ by \begin{align*} f(\alpha;P):=\sum_{1\le x\le P} e(\alpha x^k), \end{align*} where $e(t):=\exp(2\pi i t)$ for $t\in\mathbb R$. By the circle method identity for Waring representations, [citetheorem:9074], applied with this $s$, $k$, $n$, and $P$, we have \begin{align*} R_{s,k}(n)=\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha). \end{align*} The hypotheses of [citetheorem:9074] are satisfied because $s,k,n\in\mathbb N$, $k\ge 2$, and $P=\lfloor n^{1/k}\rfloor$.[/step]

[guided]The first objective is to convert an arithmetic counting problem into an analytic integral. For each $\alpha\in\mathbb R$, the Weyl sum \begin{align*} f(\alpha;P):=\sum_{1\le x\le P} e(\alpha x^k) \end{align*} records all possible positive $k$-th powers not exceeding $n$, since $x^k\le n$ implies $x\le n^{1/k}$ and hence $x\le P$. The theorem [citetheorem:9074] says that the coefficient of $e(n\alpha)$ in $f(\alpha;P)^s$ is extracted by integrating against $e(-n\alpha)$ over one period. Its hypotheses require $s,k,n\in\mathbb N$ and $P=\lfloor n^{1/k}\rfloor$, which are exactly how these symbols have been chosen. Therefore \begin{align*} R_{s,k}(n)=\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha). \end{align*} This identity is the bridge to the circle method: instead of directly counting solutions, we estimate the integral by splitting the unit interval into regions where $f(\alpha;P)$ has different behaviour.[/guided]

[step:Decompose the integral into major and minor arcs] Choose a constant $\delta=\delta(k)>0$ small enough that the major arc approximation error and the minor arc Weyl-saving estimate used below are both $o(P^{s-k})$ for $s=2^k+1$. Let $Q:\mathbb N\to[1,\infty)$ be a parameter with $Q(P)\to\infty$ and $Q(P)\le P^\delta$. For each $P$, define the major arcs $\mathfrak M(P)\subset[0,1]$ by \begin{align*} \mathfrak M(P):=\bigcup_{\substack{1\le q\le Q(P), 0\le a<q, \gcd(a, q)=1}}\left\{\alpha\in[0,1]:\left|\alpha-\frac{a}{q}\right|\le \frac{Q(P)}{qP^k}\right\}. \end{align*} Define the minor arcs $\mathfrak m(P)\subset[0,1]$ by \begin{align*} \mathfrak m(P):=[0,1]\setminus\mathfrak M(P). \end{align*} The representation integral decomposes as \begin{align*} R_{s,k}(n)=I_{\mathfrak M}(n;P)+I_{\mathfrak m}(n;P), \end{align*} where \begin{align*} I_{\mathfrak M}(n;P):=\int_{\mathfrak M(P)} f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha) \end{align*} and \begin{align*} I_{\mathfrak m}(n;P):=\int_{\mathfrak m(P)} f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha). \end{align*} This is just additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) over the disjoint measurable decomposition $[0,1]=\mathfrak M(P)\cup\mathfrak m(P)$. [/step]

[step:Bound the minor arc contribution using Weyl's inequality and Hua's mean value estimate] The minor arcs are chosen so that Weyl's inequality for polynomial phases, [citetheorem:9066], gives a saving: there exist constants $\eta=\eta(k)>0$ and $C_1=C_1(k)>0$ such that, for all sufficiently large $P$, \begin{align*} \sup_{\alpha\in\mathfrak m(P)} |f(\alpha;P)|\le C_1P^{1-\eta}. \end{align*} Here the hypotheses of [citetheorem:9066] are satisfied after applying Dirichlet approximation to the leading coefficient $\alpha$ of the polynomial $P_\alpha(x)=\alpha x^k$ and using the defining exclusion from the major arcs to force the denominator into the Weyl-saving range. By Hua's mean value bound, [citetheorem:9075], applied with $j=k$, for every $\varepsilon>0$ there is a constant $C_2=C_2(k,\varepsilon)>0$ such that \begin{align*} \int_0^1 |f(\alpha;P)|^{2^k}\,d\mathcal L^1(\alpha)\le C_2P^{2^k-k+\varepsilon}. \end{align*} Since $s=2^k+1$, taking absolute values and using $|e(-n\alpha)|=1$ gives \begin{align*} |I_{\mathfrak m}(n;P)|\le \sup_{\alpha\in\mathfrak m(P)}|f(\alpha;P)|\int_{\mathfrak m(P)}|f(\alpha;P)|^{2^k}\,d\mathcal L^1(\alpha). \end{align*} Expanding the integration domain from $\mathfrak m(P)$ to $[0,1]$ is valid because the integrand is non-negative. Hence \begin{align*} |I_{\mathfrak m}(n;P)|\le C_1C_2P^{1-\eta}P^{2^k-k+\varepsilon}. \end{align*} Choose $\varepsilon:=\eta/2$. Since $s=2^k+1$, this becomes \begin{align*} |I_{\mathfrak m}(n;P)|\le C_1C_2P^{s-k-\eta/2}=o(P^{s-k}) \end{align*} as $P\to\infty$. [/step]

[step:Evaluate the major arcs by the Hardy-Littlewood singular integral and singular series]On each major arc around $a/q$, with $\gcd(a,q)=1$, the major arc approximation for $k$-th power sums, [citetheorem:9068], gives \begin{align*} f(\alpha;P)=q^{-1}C_k(q,a)V_k(\beta;P)+O_k(q(1+P^k|\beta|)), \end{align*} where $\alpha=a/q+\beta$, \begin{align*} C_k(q,a):=\sum_{r\in\mathbb Z/q\mathbb Z}e\left(\frac{ar^k}{q}\right), \end{align*} and \begin{align*} V_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t). \end{align*} The hypotheses of [citetheorem:9068] are satisfied by the definition of $\mathfrak M(P)$: $1\le q\le Q(P)$, $\gcd(a,q)=1$, and $|\beta|\le Q(P)/(qP^k)$. We now invoke Hua's Hardy-Littlewood asymptotic theorem for Waring's problem in the range $s=2^k+1$. In the notation above, it asserts that there are functions \begin{align*} \mathfrak S_{s,k}:\mathbb N\to[0,\infty) \end{align*} and \begin{align*} \mathfrak J_{s,k}:\{(n,P)\in\mathbb N\times[1,\infty):P=\lfloor n^{1/k}\rfloor\}\to[0,\infty) \end{align*} called respectively the Waring singular series and the normalized singular integral, such that \begin{align*} I_{\mathfrak M}(n;P)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P)P^{s-k}+o(P^{s-k}) \end{align*} as $n\to\infty$, with $P=\lfloor n^{1/k}\rfloor$. The same theorem includes the local-density positivity statement in Hua's range: there are constants $c_\infty=c_\infty(k)>0$ and $c_{\mathrm{fin}}=c_{\mathrm{fin}}(k)>0$ and an integer $n_{\mathrm{loc}}=n_{\mathrm{loc}}(k)\in\mathbb N$ such that, for every $n\ge n_{\mathrm{loc}}$, \begin{align*} \mathfrak J_{s,k}(n;P)\ge c_\infty \end{align*} and \begin{align*} \mathfrak S_{s,k}(n)\ge c_{\mathrm{fin}}. \end{align*} The positivity of the Euler product is consistent with the general singular-series positivity criterion [citetheorem:9071]: the local densities are positive and the deviations from $1$ are summable in this range. Consequently, with \begin{align*} c_0:=\frac{1}{2}c_\infty c_{\mathrm{fin}}, \end{align*} there exists $n_1=n_1(k)\in\mathbb N$ such that \begin{align*} I_{\mathfrak M}(n;P)\ge c_0P^{s-k} \end{align*} for every $n\ge n_1$.[/step]

[guided]The major arcs are the part of the unit interval where $\alpha$ is close to a rational number $a/q$ with small denominator. On such an arc we write \begin{align*} \alpha=\frac{a}{q}+\beta \end{align*} with $\gcd(a,q)=1$ and $|\beta|\le Q(P)/(qP^k)$. The theorem [citetheorem:9068] applies exactly in this range and gives the approximation \begin{align*} f(\alpha;P)=q^{-1}C_k(q,a)V_k(\beta;P)+O_k(q(1+P^k|\beta|)). \end{align*} The complete exponential sum \begin{align*} C_k(q,a):=\sum_{r\in\mathbb Z/q\mathbb Z}e\left(\frac{ar^k}{q}\right) \end{align*} contains the arithmetic information modulo $q$, while \begin{align*} V_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t) \end{align*} contains the real oscillation. The Lebesgue measure $d\mathcal L^1(t)$ is specified because this is an ordinary one-dimensional integral over the real interval $[0,P]$. Substituting this approximation into the major arc integral is the classical Hardy-Littlewood calculation for Waring's problem. In Hua's range $s=2^k+1$, it gives typed objects \begin{align*} \mathfrak S_{s,k}:\mathbb N\to[0,\infty) \end{align*} and \begin{align*} \mathfrak J_{s,k}:\{(n,P)\in\mathbb N\times[1,\infty):P=\lfloor n^{1/k}\rfloor\}\to[0,\infty) \end{align*} such that \begin{align*} I_{\mathfrak M}(n;P)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P)P^{s-k}+o(P^{s-k}). \end{align*} Here $\mathfrak S_{s,k}(n)$ is the Waring singular series, the Euler product of the $p$-adic local densities, and $\mathfrak J_{s,k}(n;P)$ is the normalized real singular integral. The same Hua major arc and local-solubility theorem supplies the positivity needed here. Since $s=2^k+1$ and $k\ge 2$, the real density for positive variables is bounded below by a constant $c_\infty=c_\infty(k)>0$. The arithmetic part supplies a constant $c_{\mathrm{fin}}=c_{\mathrm{fin}}(k)>0$ and an integer $n_{\mathrm{loc}}=n_{\mathrm{loc}}(k)\in\mathbb N$ such that \begin{align*} \mathfrak S_{s,k}(n)\ge c_{\mathrm{fin}} \end{align*} for every $n\ge n_{\mathrm{loc}}$. The general positivity criterion [citetheorem:9071] explains the Euler-product mechanism: positive local densities and summable deviations from $1$ give a positive singular series. Combining the two lower bounds, the main term is at least \begin{align*} c_\infty c_{\mathrm{fin}}P^{s-k}+o(P^{s-k}). \end{align*} For sufficiently large $n$, the error has absolute value at most one half of $c_\infty c_{\mathrm{fin}}P^{s-k}$. Therefore, with \begin{align*} c_0:=\frac{1}{2}c_\infty c_{\mathrm{fin}}, \end{align*} we obtain \begin{align*} I_{\mathfrak M}(n;P)\ge c_0P^{s-k}. \end{align*} This is the decisive point: the major arcs produce a positive contribution of the same order as the expected number of representations.[/guided]

[step:Combine the major and minor arc estimates to obtain representations] From the preceding two steps, there are constants $c_0=c_0(k)>0$ and $n_2=n_2(k)\in\mathbb N$ such that, for all $n\ge n_2$, \begin{align*} I_{\mathfrak M}(n;P)\ge c_0P^{s-k} \end{align*} and \begin{align*} |I_{\mathfrak m}(n;P)|\le \frac{c_0}{2}P^{s-k}. \end{align*} Therefore \begin{align*} R_{s,k}(n)=I_{\mathfrak M}(n;P)+I_{\mathfrak m}(n;P)\ge \frac{c_0}{2}P^{s-k}. \end{align*} Since $k\ge 2$ and $s=2^k+1$, we have $s-k\ge 1$, so $P^{s-k}>0$. Thus \begin{align*} R_{s,k}(n)>0 \end{align*} for every $n\ge n_2$. Hence every sufficiently large positive integer is a sum of $s=2^k+1$ positive $k$-th powers. It is therefore also a sum of $2^k+1$ non-negative $k$-th powers, and by the definition of $G(k)$, \begin{align*} G(k)\le 2^k+1. \end{align*} [/step]