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For an interval set avoiding one residue at each sieving prime, the Ramanujan sum at the forbidden Chinese remainder theorem residue gives a linear combination of Fourier samples equal to $\mu (d)|S|$. Cauchy-Schwarz inequality gives sample energy at least $|S{|}^2/\phi (d)$ for each squarefree modulus. Summing these energies and applying the analytic large sieve inequality gives the bound. If primes dividing $q$ are excluded, restrict the denominator to $(d,q)=1$; it is uniformly at least a constant times $(\phi (q)/q)\log (2Q)$.

For $(a,q)=1$ and $q\le {\log }^{A}N$, apply the Siegel–Walfisz theorem to every reduced residue class with error exponent $A+B+2$. The main terms sum to $N{c}_q(a)/\phi (q)$, where the Ramanujan sum satisfies $c_q(a)=\mu (q)$ for coprime $a,q$. Nonunits with nonzero Von Mangoldt function are powers of primes dividing $q$ and contribute at most $\omega (q)\log N$, which is absorbed by the error. For arbitrary $a$, the correct main term is $N{c}_q(a)/\phi (q)$; the reduced numerator hypothesis cannot be omitted.

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.

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.

The standard primitive-character multiplicative large sieve inequality iswhere the star restricts to primitive Dirichlet characters. This is the form used in analytic arguments for Linnik's theorem. The prime-power Gauss identities extend to arbitrary primitive conductors by the Chinese remainder theorem. Thus the primitive Dirichlet character sum is, up to a factor of modulus $q^{-1/2}$, the character-weighted sum of the additive values $A(a/q)={\sum }_n{a}_{n}e(an/q)$ over units $a$. Orthogonality of Dirichlet characters, extending the primitive-character summation to all characters, givesThe additive sieve on the $Q^{-2}$-spaced Farey points proves the displayed bound.

For both prime-interval applications, use the following large sieve upper bound for sifted intervals. Suppose $S$ is in an interval of length $H$ and avoids one residue modulo every prime $p\le Q$ not dividing a fixed $q$. ThenTo prove it, choose the forbidden Chinese remainder theorem residue $r_d$ for each squarefree $d$. The Ramanujan sum $c_d(n-r_d)$ equals $\mu (d)$ on $S$, since $n-r_d$ is a unit modulo $d$. Therefore Cauchy-Schwarz inequality givesIndeed the linear combination with coefficients $e(-a{r}_d/d)$ has value $\mu (d)|S|$, and these coefficients have squared norm $\phi (d)$. Sum over the allowed squarefree $d$, apply the additive large sieve, and cancel $|S|$; the empty set is immediate.

Finally ${\mathcal{L}}_1(Q)\ge c\log (2Q)$. Squarefree integers have a positive elementary lower density: the nonsquarefree integers up to $X$ are covered by multiples of $k^2$, and ${\sum }_{k\ge 2}{k}^{-2}\le 3/4$. Partial summation turns this density into the harmonic lower bound. Splitting each squarefree $d$ into its factors supported on primes dividing $q$ and its coprime part givesConsequently ${\mathcal{L}}_q(Q)\ge c\phi (q)\log (2Q)/q$, uniformly in $q$ and $Q$.

Let $g$ have order $n$ in a finite group. In a complex group representation, its matrix is diagonalizable with eigenvalues ${\zeta }_j$ satisfying ${\zeta }_j^n=1$. If each $g^i$ with $\gcd (i,n)=1$ is conjugate to $g$, the character of a representation is constant on these powers. Henceusing the stated integrality of each Ramanujan sum. Therefore $\chi (g)$ is rational. It is also an algebraic integer, being a sum of roots of unity; a rational algebraic integer is an integer. Thus $\boxed{\chi (g)\in \mathbb{Z}}$.

In a symmetric group, raising a permutation to a power coprime to its order preserves each cycle length and therefore its entire cycle type. Since cycle type determines the conjugacy class, the hypothesis holds, and all character values of symmetric groups are integers.

A singular series assembles the local congruence densities in an additive counting asymptotic. For three prime summands it is ${\sum }_{q\ge 1}\mu (q)c_q(n)/\phi (q{)}^3$, with Ramanujan sum $c_q(n)$. Its Euler product vanishes for even $n$ and has a uniform positive lower bound for odd $n$. Different additive problems have different local factors, so this is a specific instance rather than a universal formula.