Waring's Problem

The Waring’s Problem asks about the estimation of \[r_{k,s}(N)\;:=\; \bigl|\{(x_1,\dots,x_s)\in\mathbb{N}^s:\; x_1^k+\cdots+x_s^k=N\}\bigr|.\]

Theorem (Hardy and Littlewood). For \(s\ge s_0(k)\), there exists \(\delta=\delta(s,k)>0\) such that \[r_{k,s}(N) = \mathfrak{S}(N)\, \Gamma\!\Bigl(1+\frac1k\Bigr)^s \Gamma\!\Bigl(\frac{s}{k}\Bigr)^{-1} N^{\frac{s}{k}-1} + O\!\Bigl(N^{\frac{s}{k}-1-\delta}\Bigr),\] where \(\Gamma(x)\) is the gamma function, \(\mathfrak{S}(N)\) is the “singular series”, and we prove this for \(s_0=2^k+1\).

By orthogonality, it can be written as \[r_{k,s}(N) = \int_0^1 \!\!\Bigl(\sum_{n=1}^{P} \mathrm{e(n^k\alpha)}\Bigr)^{\!s} \mathrm{e(-N\alpha)}\, \mathrm{d}\alpha = \int_0^1 \! F(\alpha)^s \, \mathrm{e(-N\alpha)}\, \mathrm{d}\alpha,\] where \[P=\bigl\lfloor N^{1/k}\bigr\rfloor \qquad\text{and}\qquad F(\alpha)=\sum_{n=1}^{P} \mathrm{e(n^k\alpha)}.\]

Let \(N\ge 2^k\), choose \(0<\nu<\tfrac15\). Then we construct the major arcs \(\mathfrak{M}\) and minor arcs \(\mathfrak{m}\) as follows: \[\mathfrak{M} = \bigcup_{q\le P^{\nu}} \ \bigcup_{\substack{0\le a<q\\(a,q)=1}} \Bigl\{\alpha\in[0,1]:\, \bigl|\alpha-\tfrac{a}{q}\bigr| \le \frac{1}{q\,P^{k\nu}} \Bigr\}, \qquad \mathfrak{m}=[0,1]\setminus\mathfrak{M}.\]

And it is not difficult to verify that each interval in \(\mathfrak{M}\) does not overlap. Then, we can calculate \(r_{k,s}(N)\) by evaluating the integral over the major arcs \(\mathfrak{M}\) and the minor arcs \(\mathfrak{m}\).

Now, we first estimate the integral over the minor arcs.
Theorem. Let \(k\ge 2\) and \(s\ge 2^k+1\). There exists \(\delta>0\) such that \[\int_{\mathfrak{m}} \! |F(\alpha)|^s\, \mathrm{e(-N\alpha)} \, \mathrm{d}\alpha \; \ll \; P^{\,s-k-\delta},\] where the constant depends only on \(k\) and \(s\).

Proof. By Dirichlet’s approximation, if \(\alpha\in\mathfrak{m}\) with \(Q=P^{k-\nu}\), then there exist \((a,q)\) such that \((a,q)=1\), \(P^\nu<q\le Q\), and \[\left| \alpha - \frac{a}{q} \right| \le \frac{1}{qQ} \le \frac{1}{q P^{k-\nu}}.\] By Weyl’s inequality, with \(K=2^{\,k-1}\), it yields that \[F(\alpha) \ \ll\ P^{\,1+\varepsilon} \left( \frac{1}{P} + \frac{1}{q} + \frac{q}{P^k} \right)^{1/K} \ \ll\ P^{\,1+\varepsilon - \nu/k}.\]

Lemma (Hua). For \(k\ge 2\), let \[T(\alpha) \;=\; \sum_{n=1}^N \mathrm{e(\alpha n^k)}.\] Then \[\int_0^1 |T(\alpha)|^{2^k} \, \mathrm{d}\alpha \ \ll\ N^{2^k - k + \varepsilon}.\]

Hence, by applying Hua’s lemma, we obtain \[\int_{\mathfrak{m}} |F(\alpha)|^s\, \mathrm{e(-N\alpha)} \, \mathrm{d}\alpha \ \ll\ \max_{\alpha\in\mathfrak{m}} |F(\alpha)|^{\,s-2^k} \ \int_0^1 |F(\alpha)|^{\,2^k} \, \mathrm{d}\alpha\] \[\ll\ P^{\,s - k - \delta_1},\] where \[\delta_1 = \frac{\nu(2^{\,k} - k)}{k} - \bigl(s - 2^k + 1\bigr) \varepsilon > 0.\] Then, we are done.

As for the major arcs part, we introduce the following functions \[v(\beta) = \sum_{m=1}^{N} \frac{1}{k} m^{\frac1k - 1} \, \mathrm{e(\beta m)} \quad\text{and}\quad S(q,a) = \sum_{r=1}^{q} \mathrm{e(a r^k / q)}.\]

Let \(\beta = \alpha - \frac{a}{q}\), then for \(\alpha \in \mathfrak{M}\), we have \(|\beta| \le P^{\nu-k}\) and \[\begin{aligned} F(\alpha) - \frac{S(q,a)}{q} v(\beta) &= \sum_{m=1}^{P} \mathrm{e(\frac{a m^k}{q})} \, \mathrm{e(\beta m^k)} - \frac{S(q,a)}{q} \sum_{m=1}^{N} \frac{1}{k} m^{\frac1k - 1} \mathrm{e(\beta m)}\\ &= \sum_{m=1}^{N} u(m) \, \mathrm{e(\beta m)},\end{aligned}\] where \[u(m) = \begin{cases} \mathrm{e(\frac{a m^k}{q})} - \frac{S(q,a)}{q} \frac{1}{k} m^{\frac1k - 1}, &\text{if $m$ is a $k$-th power},\\[4pt] - \frac{S(q,a)}{q} \frac{1}{k} m^{\frac1k - 1}, &\text{otherwise}. \end{cases}\] Since \(|S(q,a)| \le q\) and \(\sum_{m\le t} \frac{1}{k} m^{\frac1k - 1} = t^{1/k} + O(1)\), we obtain \[\begin{aligned} U(t) &= \sum_{m \le t}u(m)\\ &=\sum_{m \le t^{1/k}} \mathrm{e(\frac{a m^k}{q})} - \frac{S(q,a)}{q} \sum_{m \le t} \frac{1}{k} m^{\frac1k - 1}\\ &= O(q).\end{aligned}\]

By Abel’s summation, it yields that \[\sum_{m=1}^{N} u(m) \mathrm{e(\beta m)} = \mathrm{e(\beta N)} U(N) - 2\pi i \beta \int_1^{N} \mathrm{e(\beta t)} U(t) \, \mathrm{d}t \ \ll\ (1 + |\beta| N) q \ \ll\ P^{\nu}.\]

Theorem. Let \[\mathfrak{S}(N,Q) = \sum_{q \le Q} \sum_{\substack{a=1 \\ (a,q)=1}}^{q} \left( \frac{S(q,a)}{q} \right)^{\!s} \mathrm{e(-Na/q)},\] and \[J^*(N) = \int_{-P^{\nu-k}}^{P^{\nu-k}} v(\beta)^s \, \mathrm{e(-N\beta)} \, \mathrm{d}\beta.\]

\[\int_{\mathfrak{M}} F(\alpha)^s \, \mathrm{e(-N\alpha)} \, \mathrm{d}\alpha = \mathfrak{S}(N, P^{\nu}) \, J^*(N) + O\!\bigl( P^{\,s-k-\delta_2} \bigr)\] where \[\delta_2 = \frac{1 - 5\nu}{k} > 0.\]

Proof. Let \(\beta = \alpha - \frac{a}{q}\), \(V = \frac{S(q,a)}{q} v(\beta)\), and \(F = F(\alpha)\), then \[\int_{\mathfrak{M}} |F^s - V^s| \, \mathrm{d}\alpha \ \ll\ \mu(m)\, (F - V)\, \bigl(F^{s-1} + F^{s-2}V + \cdots + V^{s-1}\bigr)\] \[\ll\ P^{3\nu - k} \, P^{\,s-1+\nu} \ \ll\ P^{\,s-k-\delta_2}\] where \(\delta_2 = 1 - 5\nu > 0\).

Therefore, \[\begin{aligned} \int_{\mathfrak{M}} F(\alpha)^s \, \mathrm{e(-N\alpha)} \, \mathrm{d}\alpha &= \sum_{q\le P^\nu} \sum_{\substack{a=1 \\ (a,q)=1}}^{q} \int_{\mathfrak{M}(q,a)} V^s \, \mathrm{e(-N\alpha)} \, \mathrm{d}\alpha + O\!\bigl(P^{\,s-k-\delta_2}\bigr)\\ &= \mathfrak{S}(N,P^\nu) \, J^*(N) + O\!\bigl(P^{\,s-k-\delta_2}\bigr).\end{aligned}\] This completes the proof.

Then we estimate the singular series \(\mathfrak{S}(N,P^\nu)\) and the singular integral \(J^*(N)\), respectively.

Let \[J(N) = \int_{-\infty}^{\infty} v(\beta)^s \, \mathrm{e(-N\beta)} \, \mathrm{d}\beta,\] it follows that \[J(N) = J^*(N) + O\!\bigl(P^{\,s-k-\delta_3}\bigr)\] where \[\delta_3 = \nu \bigl( \tfrac{s}{k} - 1 \bigr) > 0.\]

Lemma. Let \(\alpha, \beta \in \mathbb{R}\) such that \(0 < \beta < 1\) and \(\alpha \ge \beta\). Then \[\sum_{m=1}^{N} m^{\beta - 1} (N-m)^{\alpha - 1} = N^{\alpha + \beta - 1} \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} + O\!\bigl(N^{\alpha - 1}\bigr),\] where the implied constant depends only on \(\beta\).

Theorem. If \(s \ge 2\), then \[J(N) = \Gamma\!\Bigl(1 + \tfrac{1}{k}\Bigr)^{\!s} \, \Gamma\!\Bigl(\tfrac{s}{k}\Bigr)^{-1} \, N^{\,\frac{s}{k} - 1} + O\!\bigl( N^{\frac{s-1}{k} - 1} \bigr).\]

Proof. Let \[J_s(N) = \int_{-\infty}^{\infty} v(\beta)^s \, \mathrm{e(-N\beta)} \, \mathrm{d}\beta\] for \(s \ge 1\). We shall prove the theorem by induction on \(s\). For \(s = 2\), we obtain \[J_2(N) = k^{-2} \sum_{m=1}^{N} m^{\frac1k - 1} (N-m)^{\frac1k - 1} = \frac{\Gamma(1+\frac1k)^2}{\Gamma(\frac{2}{k})} \, N^{\frac{2}{k} - 1} + O\!\bigl( N^{\frac1k - 1} \bigr).\] By induction, for \(s \to s+1\), we have \[\begin{aligned} J_{s+1}(N) &= \frac1k \sum_{m=1}^{N} m^{\frac1k - 1} \, J_s(N-m)\\ &= \frac{\Gamma(1+\frac1k)^{s+1}}{\Gamma\bigl( \frac{s+1}{k} \bigr)} \, N^{\frac{s+1}{k} - 1} + O\!\bigl( N^{\frac{s}{k} - 1} \bigr).\end{aligned}\] This completes the induction.

\(\mathfrak{S}(N,q)\) is not exactly the singular series that we are going to estimate; instead, we work on \[\mathfrak{S}(N) = \sum_{q=1}^{\infty} A_N(q)\] where \[A_N(q) = \sum_{\substack{a=1 \\ (a,q)=1}}^{q} \left( \frac{S(q,a)}{q} \right)^{\!s} \mathrm{e(-Na/q)}.\] By Weyl’s inequality, \[\frac{S(q,a)}{q} \ \ll\ q^{-1/k + \varepsilon}\] with \(K = 2^{k-1}\), for sufficiently small \(\varepsilon\), so that \[\delta_4 = \frac1k - s\varepsilon > 0,\] it follows that \[A_N(q) \ \ll\ \frac{q}{q^{s/k - s\varepsilon}} \ \le\ \frac{1}{q^{1+\delta_4}}.\] Hence, \(\mathfrak{S}(N)\) converges absolutely, and \[\mathfrak{S}(N) - \mathfrak{S}(N, P^\nu) \ \ll\ P^{-\nu \delta_4}.\] By the property that \(S(qr, ar + bq) = S(a,q) S(r,b)\) for \((a,r)=1\), then for \((q,r)=1\), we have \[A_N(qr) = A_N(q) \, A_N(r).\] Let \(M_N(q)\) denote the number of solutions of the congruence \[x_1^k + x_2^k + \cdots + x_s^k \equiv N \pmod{q}\] in integers \(x_i\) such that \(1 \le x_i \le q\) for \(i=1,\dots,s\). Let \(s \ge 2^k + 1\). For every prime \(p\), the series \[\chi_N(p) = 1 + \sum_{h=1}^{\infty} A_N(p^h)\] converges; then \[\mathfrak{S}(N) = \prod_{p} \chi_N(p).\]

Note.

\[\begin{aligned} M_N(q) &= \sum_{x_1=1}^{q} \cdots \sum_{x_s=1}^{q} \frac{1}{q} \sum_{a=1}^{q} \mathrm{e(\frac{a(x_1^k + \cdots + x_s^k - N)}{q})}\\ &= \frac1q \sum_{a=1}^q S(q,a)^s \, \mathrm{e(-aN/q)} = \frac1q \sum_{d \mid q} \sum_{\substack{a=1 \\ (a,q)=d}}^{q} S(q,a)^s \, \mathrm{e(-aN/q)}\\ &= \frac1q \sum_{d \mid q} \sum_{\substack{a_1=1 \\ (a_1, q/d)=1}}^{q/d} d^s S(q/d, a_1)^s \, \mathrm{e(- (a_1 N) / (q/d))}\\ &= q^{s-1} \sum_{d \mid q} A_N\!\left( \frac{q}{d} \right).\end{aligned}\]

Therefore, \[\begin{aligned} \chi_N(p) &= \lim_{h \to \infty} \left( 1 + \sum_{j=1}^h A_N(p^j) \right)\\ &= \lim_{h \to \infty} \sum_{d \mid p^h} A_N\!\left( \frac{p^h}{d} \right)\\ &= \lim_{h \to \infty} p^{h(1-s)} \, M_N(p^h).\end{aligned}\] By considering \(M_N(p^r)\) where \(r\) is a certain positive number depending only on \(k\) and smaller than \(h\), it suffices to deduce that \[\chi_N(p) > 0, \quad \text{and} \quad c_1 < \mathfrak{S}(N) < c_2,\] where \(c_1, c_2\) depends on \(k\) and \(s\). Hence, we complete the proof of the theorem.