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If $I_m$ is an integer interval inside $[1,N]$, discrete Fourier inversion gives the displayed identity. The exponential geometric sum bound gives a common envelope $H_N(\theta )\ll \min (N,\Vert \theta {\Vert }^{-1})$, whose integral is $O(\log (2N))$. At each frequency, $\widehat{1_{I_m}}(\theta )/H_N(\theta )$ is a bounded coefficient depending only on $m$, while $e(n\theta )$ depends only on $n$. A uniform bounded-coefficient Type II sum estimate therefore survives an $m$-dependent interval cutoff at a cost of one logarithm. This is particularly useful for the condition $mn\le N$.

Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the quadratic Weyl inequality is the following: if $q\ge 1$, $(a,q)=1$ and $|\alpha -a/q|\le q^{-2}$, thenThe implied constant is absolute. In particular the familiar square-root saving is $|S|\ll (N^2/q+N+q{)}^{1/2}\log (2q{)}^{1/2}$. We prove all estimates used in this statement.

First, the finite geometric series identity giveswhere $\Vert \theta \Vert$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta ))/(1-e(\theta ))$, and $|1-e(\theta )|=2|\sin \pi \theta |\ge 4\Vert \theta \Vert$. This proves the exponential geometric sum bound.

Expand the squared quadratic exponential sum by the difference $h$ between its indices. Since $(x+h{)}^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals giveFor $q\ge 16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\Vert 2ad/q\Vert \ge 1/q$. ThereforeThus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.

For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le 4q$. Summing the harmonic series, using ${\sum }_{j\le m}{j}^{-1}\le 1+\log m$, proves a block bound $O(N+q\log (2q))$. This is the required separated reciprocal-distance sum estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, soFor $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.

Extend $f$ by zero outside $[1,N]\cap \mathbb{Z}$. Here is a dyadic endpoint convention for small Type I and Type II sums. Write $D_M=[M,2M)\cap \mathbb{Z}$, with $M$ a power of two. Type I sums are $\delta$-small if, for every $M\le N^{1/100}$, dyadic $K$ with $MK\le N$, and arbitrary integer intervals $I_m\subset D_K$,Type II sums are $\delta$-small if, for all dyadic $M,K\in [N^{1/100},N^{99/100}]$ and sequences $|a_m|,|b_n|\le 1$,Zero extension handles boxes crossing the endpoint $mn=N$. Fixed choices of open or closed dyadic endpoints do not matter, but interval uniformity and uniformity in both bounded coefficient sequences do matter. These are the estimates used in the Vinogradov Type I–II method.

An interesting technical step is Fourier separation of interval cutoffs. For an integer interval $I_m$, let $\widehat{1_{I_m}}(\theta )={\sum }_{r\in I_m}e(-r\theta )$. Discrete Fourier inversion givesThe exponential geometric sum bound gives a common envelope $H_N(\theta )\ll \min (N,\Vert \theta {\Vert }^{-1})$, with ${\int }_0^1{H}_N\ll \log (2N)$. Divide the interval transform by this envelope: at each frequency it is a bounded coefficient depending on $m$, and $e(n\theta )$ is a bounded coefficient depending on $n$. The Type II sum assumption therefore bounds the interval-cutoff sum by $O(\delta N\log N)$. Choosing the outer unit phases to turn each inner sum into its modulus extends the Type I sum bound to boxes where both variables are in the Type II range.

Take $U=V=\lfloor N^{1/3}\rfloor$. With subscripts denoting truncation of arithmetic functions, the Vaughan identity iswhere $\ast$ now means Dirichlet convolution. For completeness it follows from $\mu \ast 1=ϵ$ and $\Lambda \ast 1=\log$: expand ${\mu }_{>U}\ast {\Lambda }_{>V}\ast 1$, cancel the short terms, and retain $\Lambda$. The short first term contributes $O(N^{1/3}\log N)$ to the sum against $f$.

The next two terms give Type I sums. The logarithmic term has outer coefficient $\mu (m)$, $m\le U$, and inner factor $\log n$, removed by partial summation at a cost $O(\log N)$. In the other term the outer variable is $m=cd\le UV$ with coefficientThis logarithmic coefficient bound in Vaughan identity avoids a loss from a divisor weight. On boxes with short outer variable use the hypothesis directly; on boxes where both variables are at least $N^{1/100}$ use the extended bound proved above. Boxes with inner variable less than $N^{1/100}$ contain at most $O(N^{2/3+1/100})$ pairs in total and are treated trivially. All remaining inner scales are at most $N^{99/100}$ whenever the outer scale exceeds $N^{1/100}$.

For the final convolution, group the variables asBoth variables lie between constant multiples of $N^{1/3}$ and $N^{2/3}$, well inside the Type II range. Divide $b(n)$ by $\log N$, and apply the Type II sum estimate. If the product cutoff is kept explicitly, Fourier separation of interval cutoffs costs only one extra logarithm; zero extension also permits its omission. There are $O({\log }^{2}N)$ boxes in the dyadic decomposition. The coefficient bound and, where needed, partial summation cost at most one more logarithm. These estimates give the convenient boundWith $\delta =(\log N{)}^{-20}$, each term is $o(N)$, proving the assertion. The grouping of the long term uses the nonnegativity of the Von Mangoldt function; more usual groupings instead control a divisor coefficient by a second-moment truncation. Ben Green's notes, Sections 2.3–2.4 describe that standard alternative.

Now take the sign version of the Thue–Morse sequence, $f(n)=(-1{)}^{s_2(n)}$. Concatenation of binary digits gives $f(a{2}^j+r)=f(a)f(r)$ for $0\le r<2^j$. Expanding over all choices of digits proves the Thue–Morse Fourier productPair adjacent factors. If $u=|\cos (\pi t)|$, their modulus issince the maximum on $[0,1]$ occurs at $u=1/\sqrt{3}$. Thus $|S_j(\theta )|\ll 2^{\sigma j}$ uniformly, where $\sigma =\frac{1}{2}{\log }_{2}B=2-\frac{3}{4}{\log }_{2}3<1$.

Every integer interval $J\subset [0,N]$ is a disjoint union of aligned binary blocks with at most two blocks of each length. On such a block the digit concatenation identity and the Thue–Morse Fourier product give the same bound, up to a unit phase. Summing the geometric series proves the uniform interval cancellation for binary digit parityFor a fixed $m$, the sum ${\sum }_{n\in I_m}f(mn)$ is a sum over multiples of $m$ in an integer interval $J\subset [1,N]$. The additive character orthogonality filter givesIts modulus is $O(N^{\sigma })$, independently of $m$ and the interval. Summing at most $O(N^{1/100})$ outer values proves the small Type I sums for binary digit parity:This is smaller than $(\log N{)}^{-20}$ for sufficiently large $N$. It proves the required Type I cancellation; the substantially harder Type II cancellation is not needed for this final clause.

Write $e(t)=\exp (2\pi it)$ and use $\Vert t\Vert$ for distance to the nearest integer. The Dirichlet approximation theorem, in its pigeonhole form, says that for every real $\alpha$ and integer $Q\ge 1$ there are coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$. Indeed put the $Q+1$ fractional parts of $0,\alpha ,\dots ,Q\alpha$ into $Q$ equal intervals; two lie in the same interval. Their difference gives $|q\alpha -a|\le 1/Q$, and reducing the fraction preserves the claimed bound. This includes rational $\alpha$.

For $t\ge s$, the finite geometric series has length $L=t-s+1$. When $\theta \notin \mathbb{Z}$,The last inequality uses the concavity of sine on $[0,\pi /2]$. The triangle inequality also gives $L$. At integer $\theta$ the sum has absolute value $L$, interpreting the reciprocal as infinity. Thus the precise exponential geometric sum bound isThe interval must be nonempty; for $t<s$ its sum is zero and its length should be defined as $\max (0,t-s+1)$, rather than using a potentially negative bound.

For the separated points, assume $Q\ge 2$ and $\delta >0$. An arc of length $2r\le 1$ contains at most $1+2r/\delta$ of the points, by ordering them along the arc. Points with $\Vert {\theta }_i\Vert \le 1/Q$ contribute $O(Q+{\delta }^{-1})$. In the dyadic annulus $2^j/Q<\Vert {\theta }_i\Vert \le 2^{j+1}/Q$, each summand is $O(Q{2}^{-j})$ and there are at most $1+2^{j+2}/(Q\delta )$ points. There are $O(\log (2Q))$ nonempty annuli, giving the stronger separated reciprocal-distance sumIf there is only one point the bound is immediate; otherwise $\delta \le 1/2$. The printed $\log Q$ requires a lower bound away from $Q=1$; $\log (2Q)$ avoids that endpoint defect.

A useful version of the quadratic Weyl inequality is the following, with $(a,q)=1$, $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$:For $N\ge 2$, expand the square of this quadratic exponential sum by $h=n-m$:The exponential geometric sum bound therefore gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. For $q\ge 8$, partition these $h$ into blocks of length at most $\lfloor q/4\rfloor$. Two different points in one block have difference $d$ with $0<|d|<q/4$. The reduced denominator of $2a/q$ is at least $q/2$, so $\Vert 2ad/q\Vert \ge 1/q$, whileThus the rotations $2\alpha h$ are separated by at least $1/(2q)$ in each block. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, and henceFor $q<8$ or $N<2$, the trivial bound supplies the same assertion with an absolute constant. The same proof applies to ${\sum }_{n=1}^{N}e(\alpha n^2)$, whose difference sums also have interval endpoints.

We finally prove a polynomial rate of quantitative quadratic recurrence, without using the optional finite-group assumption. For $0<\epsilon \le 1/2$, use the periodic triangular bump $b(t)=(1-\Vert t\Vert /\epsilon {)}_{+}$. It is ${\epsilon }^{-1}$ times the periodic convolution of two interval indicators, and therefore has the nonnegative Fourier coefficientsIts Fourier series is absolutely convergent, ${\sum }_r\hat{b}(r)=b(0)=1$, and the coefficient tail is ${\sum }_{|r|>R}\hat{b}(r)\ll 1/(\epsilon R)$. Suppose that $\Vert \alpha n^2\Vert \ge \epsilon$ for every $1\le n\le N$. Then ${\sum }_{n}b(\alpha n^2)=0$. Choose $R=\lceil C{\epsilon }^{-2}\rceil$ with a sufficiently large absolute $C$, so that the contribution of frequencies $|r|>R$ has absolute value at most $N\epsilon /2$. The zero frequency contributes $N\epsilon$. Since all coefficients are nonnegative and their sum is one, some $1\le r\le R$ must satisfyApply the Dirichlet approximation theorem to $r\alpha$ with $Q=N$, obtaining a reduced $a/q$, $q\le N$. The quadratic Weyl inequality then impliesSet $\epsilon =N^{-1/8}$. For sufficiently large $N$, the $N\log (2N)$ term is absorbed on the left, giving $q\ll {\epsilon }^{-2}\log (2N)$. The integer $n=rq$ consequently satisfiesMoreover the approximating integer is $arq$, andThis contradicts the assumed avoidance. Increasing the absolute constant handles the finitely many smaller $N$, uniformly for every real $\alpha$. Thus

The rational approximation below is taken in lowest terms, $(a,q)=1$, with $1\le q\le Q$, and $X>0$. This is necessary for denominator-sensitive exponential sum estimates and for the final Ramanujan sum to equal $\mu (q)$. If $0<X<1$, the defining sum over $a\le X$ is empty and $S_4=0$; hence we may assume $1\le X<\sqrt{N}$.

Set $c_u={\sum }_{a\mid u,a\le X}\mu (a)$, so $|c_u|\le \tau (u)$, and write the given Vaughan identity term as the Type II sumUse a dyadic decomposition into $u\in (U,2U]$, $d\in (D,2D]$, retaining $ud\le N$. Empty blocks may be discarded; every remaining block has $U,D\ge X$, $UD<N$, and there are $O({\log }^2(2N))$ blocks. Denote a block by $T$. The Cauchy-Schwarz inequality givesBy the allowed divisor-square summatory bound, the first factor is $O(U{\log }^3(2N))$. Expand the second factor. The diagonal $d_1=d_2$ contributes $O(UD{\log }^2(2N))$, using $0\le \Lambda (d)\le \log (2N)$. For a pair $d_1\ne d_2$, the allowed $u$ form the intervalThus the exponential geometric sum bound applies to this interval even though the original block has a hyperbolic cutoff. For each difference $h=|d_1-d_2|\le D$, there are $O(D)$ ordered pairs, giving an upper boundfor the expanded second factor, with an absolute constant $C$ accommodating integer endpoints. The supplied reciprocal fractional-part sum near a rational, with $M\asymp D$ and $R\asymp U$, bounds the last sum byOne may also derive this directly from the separated reciprocal-distance sum by splitting $h$ into blocks of length at most $q/2$; the approximation error between two block points is at most $1/(2q)$, and the rational rotations are separated by $1/q$.

Combining the estimates yieldsSince $UD<N$ and $U,D\ge X$, both $U,D\le N/X$. Taking square roots provesSumming the dyadic blocks establishes the rational-phase Type II estimate and, in particular, the requested bound with an explicit logarithmic power:

The Siegel–Walfisz theorem states that for fixed $A,C>0$, uniformly for $q\le (\log N{)}^A$ and $(b,q)=1$,The constant is allowed to be ineffective. Apply this with $C=A+B+2$ to the reduced residue classes. Their contribution to the prime exponential sum issince there are at most $q\le (\log N{)}^A$ residue classes. To evaluate the Ramanujan sum, the Möbius divisor-sum identity gives $1_{(b,q)=1}={\sum }_{d\mid (b,q)}\mu (d)$. Then the finite geometric series givesWhen $(a,q)=1$, only $m=1$ remains, giving $c_q(a)=\mu (q)$. The omitted nonunits contribute only prime powers $p^j$ with $p\mid q$. For each such prime, the sum of their Von Mangoldt function weights is at most $\log N$, so their total is at most $\omega (q)\log N=O_A(\log N\log \log (3N))$, absorbed by the stated error. This proves the major-arc value of the prime exponential sum:The original PDF does not state coprimality in this last request. Without it the asserted main term is genuinely false: $a=q=2$ makes $e(an/q)=1$ and hence ${\hat{\Lambda }}_N(a/q)=\psi (N)\sim N$, whereas $\mu (2)/\phi (2)=-1$. For an arbitrary numerator the correct main term is $N{c}_q(a)/\phi (q)$, with $c_q(a)$ given by the divisor formula above. Reducing the fraction before applying the boxed result is the equivalent repair.

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

In the fourth moment from part c, putThen the phase is $e(\alpha s_1{s}_2{u}_1{u}_2)$. The contribution with $s_1=0$ is $O(R{T}^2)$, because $r_1=r_2$; the contribution with $u_1=0$ is $O(T{R}^2)$. Their fourth roots are respectively $R^{1/4}{T}^{1/2}$ and $R^{1/2}{T}^{1/4}$.

Off the diagonals, fix $s_2,u_1,u_2$. The variable $s_1$ ranges over an interval of length $O(R)$, subject only to harmless parity and range restrictions. The exponential geometric sum bound givesPut $m=|s_2{u}_1{u}_2|$. We have $m\le CR{T}^2$, and the number of representations of a fixed $m$ by the three factors, including signs and range restrictions, is $O({\tau }_4(m))$. Hence the fourth moment isUsing $(A+B+C{)}^{1/4}\le A^{1/4}+B^{1/4}+C^{1/4}$ in part c provesThis is the factorized fourth moment for a bilinear quadratic exponential sum.

For coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$, expand the squared quadratic exponential sum by its difference $h$. The exponential geometric sum bound gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. In each block of length at most $q/4$, the rotations $2\alpha h$ are separated by at least $1/(2q)$: the reduced denominator of $2a/q$ is at least $q/2$, and the approximation error between two block points is at most $1/(2q)$. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, giving the displayed inequality. For bounded $q$ the trivial estimate supplies the same conclusion.

Suppose $|a_u|\le \tau (u)$, $|b_d|\le \log (2N)$, $u,d>X\ge 1$, $ud\le N$, and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$. A dyadic decomposition reduces $S=\sum a_u{b}_{d}e(\theta ud)$ to blocks $u\asymp U,d\asymp D$. The Cauchy-Schwarz inequality, divisor-square summatory bound and the reciprocal fractional-part sum near a rational giveThe condition $ud\le N$ makes the allowed $u$ values for each pair $d_1,d_2$ an interval, so the exponential geometric sum bound applies despite the hyperbolic cutoff. Since $UD\le N$ and $U,D\gg X$, take square roots and sum the $O({\log }^2(2N))$ blocks to get the displayed estimate.