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Translating an integer interval by $xy$ changes a sum of unit-modulus terms by at most $2xy$. Averaging gives the displayed formula. For $n\asymp N$, $Z=N^{2/5}$, expand $-t\log (1+xy/n)$ to degree $r=\lfloor 5.01\log t/\log N\rfloor$. Its remainder is at most $t{N}^{-(r+1)/5}<t^{-1/500}$. This reduces the logarithmic exponential sum to bilinear polynomial sums while keeping explicit boundary and approximation errors.

Cancellation occurs when different complex numbers in an exponential sum have directions that reduce the absolute value of their sum. For example, the orthogonality of roots of unity makes ${\sum }_{a=0}^{p-1}e(a/p)=0$ for every integer $p>1$, despite all terms having absolute value one.

If a circle-valued sequence is not an equidistributed sequence, some positive-shift difference is not equidistributed either. The contrapositive follows by applying the Van der Corput inequality for finite scalar sequences to every nonzero integer exponential sum. Thus cancellation for all fixed nonzero differences implies equidistribution of the original sequence.

If ${\theta }_r$ have circular spacing at least $\delta$, thenMultiply the exponential sum by $e(-Mt)$, apply the Sobolev–Gallagher inequality on disjoint arcs of length $\delta$, and sum. The finite-interval Parseval identities and Cauchy-Schwarz inequality bound the derivative contribution by $2\pi N\sum |a_n{|}^2$.

For fixed $c,h>0$, $L\ge 1$ and $\epsilon \ge 0$, complete the square in the exponent. A translated Gaussian function summed on a fixed-spaced lattice has mass $O(\sqrt{L})$ uniformly in the translate, by comparison with its integral and its maximum. This keeps a square-root logarithm, rather than the full number of dyadic intervals, when summing exponentially damped exponential sum bounds.

For $s=\sigma +it$, $\sigma >0$, and $x\ge |t|/\pi$, the displayed approximation holds away from the pole. Apply the Van der Corput sum-integral lemma to $-t\log w/(2\pi )$ on $w\ge x$, where its derivative has modulus at most one half. Abel summation with $w^{-\sigma }$ converts the bounded unweighted discrepancy to $O(x^{-\sigma })$. Locally uniform convergence of this weighted discrepancy extends the identity from $\sigma >1$ to $\sigma >0$. It turns estimates for finite exponential sums into estimates for the Riemann zeta function.

The circle method represents a weighted additive count as an integral of products of exponential sums against a character. Near rationals with small denominator, major arcs produce a structured main term involving a singular series and an archimedean singular integral. Away from those rationals, minor arcs require cancellation bounds. The partition and approximation estimates depend on the problem; a formal integral alone is not a positivity proof.

For $1\le j\le k$ and every $\eta >0$, the displayed estimate bounds a mean value of a polynomial exponential sum. For cubes its eighth-moment bound is $O_{\eta }(n^{5+\eta })$. It complements pointwise Weyl differencing estimates: a ninth moment over a region is bounded by its supremum there times this eighth moment. Leiden lecture notes, Section 8.1 give the cubic formulation.

The large sieve bounds how much an exponential sum can concentrate at separated points of the circle group. Its variance form of the large sieve also bounds simultaneous concentration in residue classes modulo many primes.

In a circle-method problem of scale $n$, major arcs are neighborhoods of reduced rational points $a/q$ with small denominator, where the relevant exponential sum is approximated using distribution in residue classes. For the three-prime problem one may take $q\le (\log n{)}^B$ and $|\theta -a/q|\le (\log n{)}^B/n$, with suitably fixed $B$.

For an integer $m$, direct integration gives zero unless $m=0$, in which case it gives one. Products of this identity on the unit cube turn integrals of finite exponential sums into counts of integer solutions. This is the elementary orthogonality used in the Vinogradov mean value.

Vinogradov's three-primes theorem states that every sufficiently large odd positive integer is a sum of three primes, with repetitions allowed. The parity restriction is a genuine local obstruction to the main asymptotic, not merely a technical convenience.

Let $n$ be large and define the prime-weighted exponential sumwhere $\Lambda$ is the Von Mangoldt function. Character orthogonality gives the Hardy-Littlewood circle method identityThe task is to extract a positive main term from rational neighborhoods and make the remaining integral smaller.

Take $Q=(\log n{)}^B$, with a fixed sufficiently large $B$, and use major arcs $|\theta -a/q|\le Q/n$ around reduced fractions with $q\le Q$. They are disjoint once $n$ is large relative to $Q^3$. The minor arcs are their complement. On a major arc put $\theta =a/q+\beta$. The Siegel–Walfisz theorem, followed by partial summation in each residue class, givesuniformly on these arcs after choosing the underlying logarithmic error exponent sufficiently large. The reduced residue-class sum is the Ramanujan sum $c_q(a)=\mu (q)$ when $(a,q)=1$. Prime powers dividing $q$ and the partial-summation factor $1+n|\beta |$ are absorbed in the stated error choice.

Cubing and integrating produces the singular series and singular integral:The last identity is Fourier inversion for the volume of the simplex $t_1+t_2+t_3=n$, $t_i\ge 0$; the upper bounds $t_i\le n$ do not restrict its interior. The tails and approximation errors are $o(n^2)$. Since $\mu (q{)}^3=\mu (q)$ on squarefree $q$, multiplicativity givesFor even $n$ the factor at $2$ vanishes. For odd $n$ it is $2$, andThe lower bound is uniform in odd $n$. The product converges because the sums of the local deviations converge. It describes how congruence restrictions alter the otherwise random prime-summand density.

For the minor arcs, a Type I/II prime-sum estimate provides the necessary cancellation. One useful form, obtainable from Vaughan's identity and bilinear estimates, iswhen $|\theta -a/q|\le q^{-2}$. Dirichlet's approximation theorem supplies a denominator $q\le n/Q$ with $|\theta -a/q|\le Q/(qn)$. On the minor arcs its denominator must exceed $Q$, so both denominator-dependent terms are at most $n/\sqrt{Q}$. Choosing $B$ large gives ${\sup }_{\mathfrak{m}}|S(\theta )|=o(n/\log n)$. The mechanism is to split $\Lambda$ into short-divisor Type I sums and bilinear Type II sums; geometric-sum bounds treat Type I, while Cauchy-Schwarz exposes correlations and rational spacing in Type II. The displayed estimate is the substantive analytic input, rather than a consequence of bounding each summand separately.

Meanwhile Parseval identity and the elementary prime-weight second moment giveThereforeCombining the two regions yields $R_{\Lambda }(n)=\frac{1}{2}\mathfrak{S}(n)n^2+o(n^2)$. Proper prime powers account for only $O(n^{3/2}(\log n{)}^4)$: there are $O(\sqrt{n}\log n)$ such choices for one coordinate, at most $n$ choices for another, and each weight is at most $\log n$. Hence they cannot account for the positive main term. The weighted count using primes alone is positive for every sufficiently large odd $n$, proving the asserted representation. The conclusion is the sufficiently-large three-prime theorem; no verification of every small odd integer is supplied by this argument.

Use the integer norm lower bound for rational approximation to a quadratic irrational. The integer $2q^2-a^2$ is nonzero, since $\sqrt{2}$ is irrational. If $a/q<2$, thenbecause $\sqrt{2}+a/q<\sqrt{2}+2<4$. If $a/q\ge 2$, the difference is at least $2-\sqrt{2}>1/4\ge 1/(4q^2)$. ThusIn particular, $\sqrt{2}$ is a badly approximable number.

To estimate the prime-weighted exponential sum, let $Q=\lfloor N^{1/2}\rfloor$. The Dirichlet approximation theorem gives a reduced fraction $a/q$ withCombining this with the lower bound gives $q>Q/4$. Thus $q\asymp N^{1/2}$; the approximation cannot have a small denominator.

The relevant analytic input is the prime exponential-sum estimate from Vaughan identity:Here is the structure behind that estimate. With Dirichlet convolution and cutoffs $U,V$, the Vaughan identity isIt follows from $\Lambda =\mu \ast \log$, $\log =\Lambda \ast 1$, and $\mu \ast 1={\delta }_1$. The first two Dirichlet convolution terms become type I sums with one short factor; the last becomes a bilinear type II sum with both factors bounded below. Split these sums into dyadic blocks and use $U=V=N^{2/5}$. For the type I sums, the inner finite geometric series has bound $\min (L,(2\Vert \alpha m\Vert {)}^{-1})$. For the type II sums, the Cauchy-Schwarz inequality followed by expansion introduces analogous sums in differences of indices. Rational approximation separates the phases into blocks of length comparable to $q$, bounding these reciprocal-distance sums. The Dirichlet convolution coefficients are divisor-bounded, producing the logarithmic factors. These steps give the displayed estimate; no unproved cancellation of arbitrary weighted sums is being assumed.

With $\alpha =\sqrt{2}$ and $q\asymp N^{1/2}$, the two denominator-dependent terms are $O(N^{3/4})$. ConsequentlyThe final passage absorbs the logarithm using the positive exponent margin $9/10-4/5=1/10$.

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.

Write $e(t)=\exp (2\pi it)$. A sequence in the circle group is an equidistributed sequence if, for every interval $I\subseteq \mathbb{R}/\mathbb{Z}$,where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. Trigonometric polynomials approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant additive characters have integral zero. These facts give the Weyl criterion:This is the link between equidistribution and cancellation in exponential sums.

Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne 0$ and set $z_n=e(m\varphi (n))$. For every fixed $h\ge 1$, the Weyl criterion would giveThe omitted final $h$ terms change a normalized average by at most $h/N$.

Here is the needed Van der Corput inequality for finite scalar sequences. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to $H{\sum }_n{z}_n={\sum }_t{\sum }_{j=0}^{H-1}{z}_{t-j}$ and expand the squared inner sum. Taking $N\to \infty$ first leaves a bound $1/H$; then let $H\to \infty$. Every nonzero Fourier average of $\varphi$ vanishes, so the Weyl criterion makes $\varphi$ equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive $h$, hence for some $h\ne 0$ as requested.

For $\varphi (n)=\sqrt{2}{n}^2$, the difference is ${\Delta }_h\varphi (n)=-2h\sqrt{2}n-h^2\sqrt{2}$. For any nonzero integer $m$, its exponential sum is a constant phase times a geometric progression with ratio $e(-2mh\sqrt{2})\ne 1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt{2})|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

For $0<\epsilon \le 1/2$, this continuous function on $\mathbb{R}/\mathbb{Z}$ has nonnegative Fourier coefficientsIt is ${\epsilon }^{-1}$ times the periodic convolution of two indicators of $[-\epsilon /2,\epsilon /2]$. Its Fourier series is absolutely convergent, ${\sum }_r{\hat{b}}_{\epsilon }(r)=b_{\epsilon }(0)=1$, and ${\sum }_{|r|>R}{\hat{b}}_{\epsilon }(r)\ll 1/(\epsilon R)$. Consequently, if all $N$ points $t_n$ avoid the interval $\Vert t\Vert <\epsilon$, some frequency $1\le r\ll {\epsilon }^{-2}$ satisfies $|{\sum }_{n}e(r{t}_n)|\gg N\epsilon$. This detects missing small neighborhoods by a large exponential sum.

For large positive $t$ and $0<\sigma \le 1$, a fixed sufficiently large $C$ gives the displayed upper bound; to the right of one use $|\zeta (\sigma +it)|\ll {\log }^{2/3}t$. Its nonlinear dependence on $1-\sigma$ permits a wider zero-free region of the Riemann zeta function through the Landau zero-free-region theorem. Its proof uses estimates for exponential sums; its use as an upper-bound input is separate from a proof of a zero-free region.

An integer interval is a disjoint union of aligned binary blocks with at most two of each length. On $[a{2}^j,(a+1)2^j)$ the Thue–Morse sequence sign factors as $f(a)f(r)$, so the Thue–Morse Fourier product bounds its exponential sum by $O(2^{\sigma j})$. Sum the geometric series over block lengths to obtain the uniform bound, for $\sigma =2-\frac{3}{4}{\log }_{2}3$.

Every sufficiently large odd integer is a sum of three primes, with repetitions allowed. The weighted count is asymptotic to $\mathfrak{S}(n)n^2/2$. Siegel–Walfisz theorem controls the major arcs, and Type I/II estimates for the prime exponential sum make the minor arcs negligible. Removing prime-power contributions gives genuine prime representations. This is the sufficiently-large theorem, not by itself a proof for every small odd integer.