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For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves

The Hecke operators organize modular forms simultaneously by their transformation law and their Fourier coefficients. Work with $\Gamma =S{L}_2(\mathbb{Z})$, $M_k$ its space of weight-$k$ modular forms, and $S_k$ its cusp form subspace, with even $k\ge 4$. Odd weights give zero spaces. The normalization below makes the first coefficient of $T_{n}f$ equal to the $n$th coefficient of $f$, which is the key link between operators and arithmetic.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with positive determinant, the determinant-normalized slash operator for modular forms isThe automorphy factor identity proves $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Positive scalar matrices act trivially. Let ${\mathcal{A}}_n$ be the integral matrices of determinant $n>0$. Left multiplication by $\Gamma$ reduces their first column, by Bezout identity, to $(a,0)$ with $a>0$; the determinant then gives $d=n/a>0$. A row shear makes $0\le b<d$. These are unique determinant-n matrix representatives for Hecke operators, so defineRight multiplication by any element of $\Gamma$ permutes these left cosets. The slash operator for modular forms identity therefore proves that $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic on the upper half-plane. Its cusp behavior follows explicitly from the coefficient calculation below, so $T_n$ preserves both $M_k$ and $S_k$.

Insert $f={\sum }_{r\ge 0}{a}_r{q}^r$ in the sum. The finite sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Writing $r=d\ell$ leaves the term $a_{d\ell }{q}^{a\ell }$ with multiplier $n^{k-1}{d}^{1-k}=a^{k-1}$. Thus the Fourier coefficients of a composite-index Hecke operator areFor $m=0$ use $\gcd (0,n)=n$. In particular the constant coefficient becomes ${\sigma }_{k-1}(n)a_0$, so cusp forms remain cusp forms, andfor a prime number $p$. The absence of negative powers proves cusp holomorphy. Notice that no primality assumption is needed in the composite-index formula.

The Hecke multiplication relations can be derived directly, rather than assumed from the coset description. On Fourier series define auxiliary maps $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. They need not separately preserve $M_k$, but $U_p{V}_p=1$. With $c=p^{k-1}$, the coefficient formula givesMultiplying on the left by $U_p+c{V}_p$, the term $j=0$ in the $U_p$ sum contributes $U_p^{r+1}$; the terms with $j\ge 1$ become $c^j{V}_p^{j-1}{U}_p^{r-j}$. Together with the $c{V}_p$ sum this is exactly $T_{p^{r+1}}+c{T}_{p^{r-1}}$. Hence the Hecke prime-power recurrence from q-shift operators isFor coprime $m,n$, substituting the coefficient formula twice gives a sum over $a\mid \gcd (m,r)$ and $b\mid \gcd (n,r)$ with coefficient $(ab{)}^{k-1}{a}_{mnr/(ab{)}^2}(f)$. The divisors $a,b$ correspond uniquely to a divisor $ab\mid \gcd (mn,r)$, so $T_m{T}_n=T_{mn}$. The prime-power recurrence, by induction on the smaller exponent, givesCombining the coprime and prime-power formulas provesIn particular the Hecke operators commute, and the modular Hecke algebra is generated by the prime operators.

The analytic reason for simultaneous diagonalization is the Petersson inner productwhich is positive definite on $S_k$ and convergent because of exponential cusp decay. Here is the adjoint argument. For a matrix $A$, set $H_A=\Gamma \cap A^{-1}\Gamma A$. The left cosets in $\Gamma A\Gamma$ are represented by $A\gamma$ with $\gamma \in H_A\backslash \Gamma$. Unfolding this finite coset sum in the Petersson inner product givesUnder $z=A\tau$, hyperbolic area is unchanged and the determinant factors in the slash operator for modular forms giveThe integration group becomes $A{H}_A{A}^{-1}=\Gamma \cap A\Gamma A^{-1}$. Folding this finite cover back to $\Gamma \backslash \mathbb{H}$ identifies the adjoint with the correspondence for $A^{-1}$. For $A\in {\mathcal{A}}_n$, the matrix $\mathrm{adj}(A)=n{A}^{-1}$ is again integral with determinant $n$; scalar matrices act trivially, and adjugation permutes the double cosets making up ${\mathcal{A}}_n$. The real normalizing factor $n^{k/2-1}$ is unchanged. Thus this proves that Hecke operators are self-adjoint for the Petersson inner product:The valence formula for the modular group makes $S_k$ a finite-dimensional vector space. Commuting self-adjoint operators on a finite-dimensional inner product space have a common orthogonal eigenbasis: diagonalize one operator, use commutativity to preserve each of its eigenspaces, and repeat within those spaces. Therefore $S_k$ has a basis of Hecke eigenforms, with real Hecke operator eigenvalues.

If $f$ is a common Hecke eigenform, say $T_{n}f={\lambda }_{n}f$, its first Fourier coefficient cannot vanish: $a_1=0$ would imply $a_n=a_1(T_{n}f)={\lambda }_n{a}_1=0$ for every $n$. Normalize $a_1=1$. Then the Hecke operator eigenvalue ${\lambda }_n$ is precisely the Fourier coefficient $a_n$. The Hecke multiplication relations becomeConsequently the coefficients are multiplicative at coprime indices, and their local generating function satisfiesThis gives the Euler product of a Hecke eigenformin a half-plane of absolute convergence. The Fourier coefficient bound for a cusp form already supplies such a half-plane; Question 4 improves it to $\mathrm{Re}s>(k+1)/2$.

The L-function of a cusp form also has an analytic description independent of the Hecke eigenform condition. Termwise Mellin transform in an initial right half-plane givesAt infinity the cusp form decays exponentially. Modular inversion gives $f(i/y)=(iy{)}^{k}f(iy)$, so at zero it is $O(y^{-k}{e}^{-2\pi /y})$. The integral therefore converges locally uniformly for every $s$, continuing $\Lambda$ to an entire function. Changing $y$ to $1/y$ provesFor a normalized Hecke eigenform, the Euler product of a Hecke eigenform and this functional equation describe the same completed L-function of a cusp form.

There is also an integral arithmetic structure. One standard consequence of the level-one dimension formula and the valence formula for the modular group is the integral triangular basis of level-one modular forms: if $k=12m+s$ with $s\in \{0,4,6,8,10,14\}$, choose $F_s=1,E_4,E_6,E_4^2,E_4{E}_6,E_4^2{E}_6$ respectively. Thenform a basis of $M_k$, and the terms $1\le j\le m$ form a basis of $S_k$. Their first nonzero terms are $q^j$ with coefficient one and all coefficients are integers. Successive elimination of coefficients of $q,q^2,\dots ,q^m$ shows that these cusp basis elements generate exactly the lattice of rational cusp forms with integral Fourier coefficients. The composite-index coefficient formula preserves this lattice, so each $T_n$ has an integral matrix on it. Its characteristic polynomial is monic with integral coefficients. This proves that Hecke eigenvalues are algebraic integers; self-adjointness also makes every root of that characteristic polynomial real. The field generated by the eigenvalues is a finite totally real number field: a finite set of operators spans the finite-dimensional rational modular Hecke algebra, and every conjugate of each eigenvalue remains a root of its real-rooted characteristic polynomial.

Two basic examples connect the operator theory to familiar modular forms. The normalized Eisenstein series $E_k$ has coefficients proportional to ${\sigma }_{k-1}(n)$. At a prime number its coefficient formula useswhere the second term is zero unless $p\mid m$. This follows by writing $m=p^{r}u$ with $p\nmid u$ and summing a finite geometric series. It proves $T_p{E}_k=(1+p^{k-1})E_k$, and the Hecke multiplication relations give $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Meanwhile $S_{12}=\mathbb{C}\Delta$ implies $T_n\Delta =\tau (n)\Delta$, because $a_1(\Delta )=1$. Thus the Ramanujan tau function is multiplicative and satisfies $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$. For example, $\tau (2)=-24$ gives $\tau (4)=(-24{)}^2-2^{11}=-1472$. This illustrates how Hecke operators turn analytic Fourier coefficients into a coherent arithmetic system.

Use the primitive real-analytic Eisenstein series with exactly the normalization in the PDF. Its coset expression uses ${\Gamma }_{\infty }=\{\pm \left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right):n\in \mathbb{Z}\}$, so primitive bottom rows differing by sign represent the same coset. The factor $1/2$ in the primitive-pair sum accounts for this sign identification. Unlike the sum over all nonzero integer pairs, this Eisenstein series has no additional factor $2\zeta (2w)$. Write $w$ for the Rankin–Selberg integral parameter and $s$ for the Dirichlet series parameter to keep their shift visible.

First establish an initial region where all rearrangements are absolute. The invariant norm of a modular form $y^{k/2}|f(\tau )|$ is bounded: it is $\Gamma$-invariant, continuous on a truncated fundamental domain, and tends to zero at the cusp. Integrating one period of the Fourier series givesWith $y=1/n$, this proves the Fourier coefficient bound for a cusp form $a_n=O(n^{k/2})$, and similarly $b_n=O(n^{k/2})$. Hence $D(f,g,s)$ is initially absolutely convergent for $\mathrm{Re}s>k+1$.

For $\mathrm{Re}w>1$, the primitive real-analytic Eisenstein series is locally uniformly convergent and holomorphic in $w$. On the standard fundamental domain it has at most polynomial growth in $y$, locally uniformly in this half-plane. These are the analytic properties of the Eisenstein series needed here; they can also be checked directly from its defining sum. For example, for real $\sigma >1$ and $y\ge 1$, majorize the primitive sum by the sum over all nonzero pairs and useThis bound follows by comparing a shifted one-dimensional sum with its integral and a largest summand. Summing over $m\ne 0$ gives $E(\tau ,\sigma )=O(y^{\sigma }+y^{1-\sigma })$. For complex $w$, $|E(\tau ,w)|\le E(\tau ,\mathrm{Re}w)$, and the estimates are uniform on compact parameter sets. Since $f$ and $g$ decay exponentially on the cusp end, the integral $I(f,g,w)$ is therefore holomorphic throughout $\mathrm{Re}w>1$ by dominated integration.

The product $f(\tau )\overline{g(\tau )}{y}^k$ is $\Gamma$-invariant: the two modular factors cancel the transformation $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$. With the hyperbolic area $d\mu =dxdy/y^2$, Rankin–Selberg unfolding gives, initially for $\mathrm{Re}w>2$,Absolute unfolding is justified by replacing $f\overline{g}$ by $|fg|$ and using the same finite fundamental-domain integral with $E(\tau ,\mathrm{Re}w)$. The subsequent termwise integrations can be justified particularly economically by Cauchy-Schwarz inequality: on each horizontal line,The coefficient bounds make the right side $O(y^{-k-1})$ as $y\downarrow 0$, and it decays exponentially as $y\to \infty$. Multiplication by $y^{\mathrm{Re}w+k-2}$ is integrable at zero when $\mathrm{Re}w>2$. Thus the integral of the inner Fourier series may be evaluated first. Orthogonality of the exponentials givesThe Mellin transform of each exponential isIn the same initial region the sum of the absolute values of these integrals is finite by $a_n\overline{b_n}=O(n^k)$. Consequently the exact Rankin–Selberg integral identity isIn particular the two parameters are not equal: the Dirichlet series argument is shifted by $k-1$.

The holomorphy of $I$ on $\mathrm{Re}w>1$ now gives the required continuation:The right side agrees with the original Dirichlet series on $\mathrm{Re}s>k+1$ and is holomorphic on the larger half-plane. No continuation of the Eisenstein series beyond $\mathrm{Re}w>1$ is needed.

It is important to supply a further argument for absolute convergence: analytic continuation alone does not imply it. For real $w>1$, both $E(\tau ,w)$ and $|f(\tau ){|}^2$ are nonnegative, and their fundamental-domain integral is finite by the cusp estimates above. Tonelli theorem therefore unfolds $I(f,f,w)$ even before knowing convergence of the Dirichlet series. On each fixed horizontal line, the Fourier series converges absolutely and uniformly, so integration in $x$ givesA second use of Tonelli theorem, now in $y$, proves the positive unfolding of a cusp-form square:Since every real $s>k$ has $s=w+k-1$ with $w>1$, we have absolute convergence of $D(f,f,s)$ for $\mathrm{Re}s>k$. Absolute convergence for complex $s$ follows by taking its real part. In fact Cauchy-Schwarz inequality also gives absolute convergence of $D(f,g,s)$ there from the two corresponding square sums.

Finally let $\sigma >(k+1)/2$ and choose $k<\beta <2\sigma -1$. The preceding square-coefficient bound and Cauchy-Schwarz inequality giveThis proves absolute convergence of cusp-form L-series from square coefficients, and henceEquivalently, the suggested elementary inequality works by choosing $k-\sigma <\alpha <\sigma -1$: after multiplication by $n^{-\sigma }$ its two bounds sum to a convergent $\sum n^{-(\sigma -\alpha )}$ and a convergent $\sum |a_n{|}^2{n}^{-(\sigma +\alpha )}$. The existence of this interval is precisely $2\sigma >k+1$.

In the finite-dimensional complex semisimple Lie algebra setting implicit in this root-system construction, the given Cartan subalgebra and root spaces determine a reduced crystallographic root system $R$. The ordering divides $R$ into positive roots and their negatives. The simple roots are the positive roots that cannot be written as a sum of two positive roots; a standard root-system fact is that they form a basis and that every positive root is a nonnegative integral combination of them.

Index the simple roots as ${\alpha }_1,\dots ,{\alpha }_{\ell }$ and form the Cartan matrixFor $i\ne j$, these entries are nonpositive integers and $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Construct one vertex per simple root and join distinct vertices by $A_{ij}{A}_{ji}$ lines. If root lengths differ, direct the arrow toward the shorter simple root. This labelled and oriented graph is the Dynkin diagram. Its unoriented Coxeter graph is admissible: normalizing the simple roots to unit length gives $4(v_i,v_j{)}^2=A_{ij}{A}_{ji}$. A general Lie algebra without the semisimple root-system hypotheses need not produce such a diagram.

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so

The ideal class group of a number field is the group of nonzero fractional ideals of its ring of integers under multiplication, modulo the subgroup of principal fractional ideals. For degree $n$ and $r_2$ pairs of complex embeddings, the Minkowski bound for ideal classes states that each class contains an integral ideal $I$ withThere are only finitely many integral ideals of bounded ideal norm: as sublattices of the free abelian group of integers, they have bounded index, and finitely many subgroups have any fixed index. Thus the ideal class group is finite.

The linear acoustic momentum and continuity equations are $\rho u_t=-p_x$ and $p_t=-\rho c^2{u}_x$. For a right-moving plane wave depending on $x-ct$, integration gives $\boxed{p=\rho cu}$, with zero perturbation constants. For a left-moving wave the sign is reversed. The factor $Z=\rho c$ is the acoustic impedance. Proportionality applies to a traveling wave; a superposition of opposite directions generally has no single such constant.

Use time dependence $e^{-i\omega t}$ and set $Z_j={\rho }_j{c}_j$. In the outer regions write pressures $I{e}^{i{k}_{0}x}+R{e}^{-i{k}_{0}x}$ and $TI{e}^{i{k}_{0}x}$, taking $I=1$ for convenience. The corresponding velocities are $(e^{i{k}_{0}x}-R{e}^{-i{k}_{0}x})/Z_0$ and $T{e}^{i{k}_{0}x}/Z_0$. Pressure and normal velocity are continuous at both layer boundaries. Within the cooled layer a decomposition into right and left waves gives, with $\delta =k_{1}L$,This matrix follows by writing $p=A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}$ and $Z_{1}u=A{e}^{i{k}_{1}x}-B{e}^{-i{k}_{1}x}$. Applying its inverse and using $p(0)+Z_{0}u(0)=2$ eliminates $R$, givingConsequentlyFor $\lambda \gg 1$, transmission is weak except near $k_{1}L=n\pi$, where $|T|=1$. The cooled layer acts as a resonant filter: an integer number of half-wavelengths fits within it at each resonance. The half-power width in $\delta$ is asymptotically $2/\lambda$ on either side. Since the two outer impedances agree, the transmitted power fraction is $|T{|}^2$.

Use coordinates $(i,j)\in \{0,\dots ,7{\}}^2$, with the starting corner $(0,0)$. The bishop stays on the 32 squares where $i+j$ is an even number. Its moves form a random walk on a graph, with an edge between distinct squares whenever $|i-i^{\prime }|=|j-j^{\prime }|$. If $d(i,j)$ is the number of reachable squares, the transition probability to each neighbor is $1/d(i,j)$.

The graph is connected: from $(i,j)$ move along its antidiagonal to $((i+j)/2,(i+j)/2)$, then along the main diagonal to the corner. The midpoint coordinates are integers because $i+j$ is an even number, and this square always lies on the board. Thus this is the full communicating state space of the bishop random walk.

The invariant distribution is proportional to degree. Its normalizing constant can be counted by summing the ordered pairs along each diagonal. The diagonals of one orientation have lengths $8,6,6,4,4,2,2$, contributing $144$; the other have lengths $1,3,5,7,7,5,3,1$, contributing $136$. ThereforeFor any neighboring states $s,t$, $\pi (s)P(s,t)=1/280=\pi (t)P(t,s)$, proving the detailed balance equations and hence that the chain is a reversible Markov chain.

The corner has seven neighbors. Kac's lemma gives its expected first positive return time:

Write $e(t)=e^{2\pi it}$. We first prove the needed cancellation of quadratic exponential sums, rather than invoke equidistribution without proof. The number $\sqrt{2}$ is irrational: writing it as a reduced fraction would make its numerator even from $u^2=2v^2$, and then its denominator even as well, a contradiction. Thus $2hr\sqrt{2}$ is irrational for nonzero integers $h,r$.

Fix $h\ne 0$ and set $z_n=e(h\sqrt{2}{n}^2)$ for $1\le n\le L$, extending it by zero otherwise. For integers $L\ge H\ge 1$, each term appears exactly $H$ times in the shifted sum, soFor any complex numbers $w_1,\dots ,w_J$, nonnegativity of ${\sum }_j|w_j-J^{-1}{\sum }_i{w}_i{|}^2$ gives $|{\sum }_j{w}_j{|}^2\le J{\sum }_j|w_j{|}^2$. Apply this special Cauchy-Schwarz inequality to the displayed outer sum, and then expand the inner squared modulus. This givesFor fixed $H$ and large $L$, each correlation is the geometric serieswhose absolute value is at most $2/|1-e(2hr\sqrt{2})|$, independently of $L$. This follows by multiplying the geometric sum by $1-e(2hr\sqrt{2})$; its denominator is nonzero by irrationality. Divide the inequality by $L^2$ and let $L\to \infty$, keeping $H$ fixed. It gives a limit superior at most $1/H$. Taking arbitrarily large $H$ provesThis is the quadratic instance of the Van der Corput inequality for finite scalar sequences, with its proof supplied here.

To turn cancellation into the particular approximation, put $\epsilon =10^{-8}$ and choose an integer $M>2/{\sin }^2(\pi \epsilon )$. The Fejér kernel is the explicit finite trigonometric polynomialThe expansion follows by counting pairs of indices with difference $h$. Its constant coefficient is one, so the cancellation just proved givesIf $\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \epsilon$, the geometric-series formula givesFor all sufficiently large $L$, the displayed average exceeds $1/2$. Consequently at least one $n\in \{1,\dots ,L\}$ satisfiesIt is a positive integer, as required. Every auxiliary cancellation and kernel identity used in this existence proof has been derived; a smooth cutoff or an unproved Weyl criterion is unnecessary.

A Freiman 2-isomorphism between subsets of abelian groups is a bijection $\varphi$ such thatBoth directions matter: an injective homomorphism on individual elements may still create new pair-sum relations. More generally a Freiman s-isomorphism preserves equalities of sums of $s$ elements in both directions. For a nonempty finite set, its doubling constant is $\sigma [A]=|A+A|/|A|$. A Freiman 2-isomorphism preserves this constant, since $a+b\mapsto \varphi (a)+\varphi (b)$ is a well-defined bijection of the two sumsets.

First let $A\subseteq \mathbb{Z}$ have $n$ elements and doubling at most $K$. The permitted Plünnecke-Ruzsa inequality gives $|2A-2A|\le K^{4}n$. We will obtain the half-size prime cyclic Freiman model for every prime $p>4K^{4}n$. Choose an auxiliary prime $q>p$ larger than the absolute value of every member of $2A-2A$. Such a prime exists: for any integer bound $M$, every prime divisor of $M!+1$ exceeds $M$.

Choose $\lambda$ uniformly from the nonzero residues modulo $q$. For each nonzero $d\in 2A-2A$, the residue $\lambda d$ is uniform among the nonzero residues. PutThere are at most $2(q-1)/p$ such residues, and therefore the probability that any of the nonzero differences is dilated into $T$ is at mostThe first cardinal estimate is an upper bound even if some listed residues coincide; the slightly looser probability bound is convenient. Thus there exists $\lambda$ with no such bad difference.

For this $\lambda$, let $t_a\in \{0,\dots ,q-1\}$ represent $\lambda a$. Split these representatives into the two intervals $[0,(q-1)/2]$ and $[(q+1)/2,q-1]$. Take $A^{\prime }$ from the more populated interval, so $|A^{\prime }|\ge n/2$. For any four elements of $A^{\prime }$, put $D=t_{a_1}+t_{a_2}-t_{a_3}-t_{a_4}$. Their common half-interval gives $|D|<q$, andIf the original difference is zero, then $D$ is a multiple of $q$ of absolute value less than $q$, so $D=0$. If the original difference is nonzero, $D$ is nonzero, and it cannot be a multiple of $p$: otherwise its residue belongs to $T$, contrary to the choice of $\lambda$. Therefore $a\mapsto t_a(\mathrm{mod}p)$ preserves pair-sum relations in both directions. It is injective as well, by applying the relation test to $a+a_0=b+a_0$ with any fixed $a_0\in A^{\prime }$. HenceThis proves the stated polynomial bound, including the prescribed prime rather than just an unspecified cyclic modulus.