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Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.

The resolution principle for sheaf cohomology states that an exact sequencewith $H^q(X,{\mathcal{A}}^j)=0$ for every $q>0$ computes $H^i(X,\mathcal{F})$ as the cohomology of $\Gamma (X,{\mathcal{A}}^{\bullet })$. In particular one may use a flasque resolution.

An affine morphism $\varphi :X\to Y$ is one for which the inverse image of every affine open subset of $Y$ is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Each direct image sheaf ${\varphi }_{\ast }{\mathcal{I}}^j$ is flasque, because restriction on $V\subseteq U\subseteq Y$ is restriction on ${\varphi }^{-1}V\subseteq {\varphi }^{-1}U$. On an affine open $U\subseteq Y$, the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is $\Gamma ({\varphi }^{-1}U,\mathcal{F})$. Checking exactness on this affine basis shows that ${\varphi }_{\ast }{\mathcal{I}}^{\bullet }$ is a flasque resolution of ${\varphi }_{\ast }\mathcal{F}$. The global-section complexes are identical, proving cohomology under an affine morphism:

For projective space, put $S=k[x_0,\dots ,x_n]$ with its usual grading and $U_i=\{x_i\ne 0\}$. Construct the twisting sheaf on projective space by setting its sections on $U_i$ to be the degree-$m$ part $(S_{x_i}{)}_m$, viewed as a module over $(S_{x_i}{)}_0$. It is free of rank one with generator $x_i^m$, even when $m$ is negative. On overlaps the generators differ by the unit $(x_i/x_j{)}^m$, so these modules glue to a line bundle $\mathcal{O}(m)$.

Assume first $n\ge 1$. These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with ${\bigcap }_i(S_{x_i}{)}_m$. Since $S$ is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every $S_{x_i}$ has no nonconstant denominator. Thus the intersection is $S_m$. Counting monomials by stars and bars givesHere $h_n^i(m)={\dim }_k{H}^i({\mathbf{P}}^n,\mathcal{O}(m))$. The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For $n=0$, every twist on the single point ${\mathbf{P}}^0$ is trivial, so $h_0^0(m)=1$ for every $m$ and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for $n\ge 1$.

The standard $n+1$ affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree $n$. In particular $H^i({\mathbf{P}}^n,\mathcal{O}(m))=0$ for $i>n$.

Let $H=\{x_n=0\}\cong {\mathbf{P}}^{n-1}$, with inclusion $i$. Multiplication by $x_n$ gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to $H$. Since a closed immersion is an affine morphism, the first part identifies the cohomology of $i_{\ast }{\mathcal{O}}_H(m)$ with that on $H$.

For $n=1$, the long exact sequence in sheaf cohomology and $H\cong {\mathbf{P}}^0$ giveStarting from the given vanishing for sufficiently large $m$ and descending gives $h_1^1(m)=0$ for $m\ge -1$ and $h_1^1(m)=-m-1$ for $m\le -2$.

Now let $n\ge 2$, and assume the formulas in dimension $n-1$. Restrictionis surjective for every $m$: for $m\ge 0$ it is polynomial restriction, and for $m<0$ both groups are zero since $\dim H\ge 1$. The long exact sequence in sheaf cohomology therefore makes $H^1({\mathbf{P}}^n,\mathcal{O}(m-1))\to H^1({\mathbf{P}}^n,\mathcal{O}(m))$ injective. For $2\le j<n$, dimension induction gives $H^{j-1}(H,{\mathcal{O}}_H(m))=0$, so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves $H^j({\mathbf{P}}^n,\mathcal{O}(m))=0$ for every $m$ and $0<j<n$.

The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on $H$. Hence $h_n^n(m-1)=h_n^n(m)+h_{n-1}^{n-1}(m)$. Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for $m\ge -n$ it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is ${\omega }_{{\mathbf{P}}^n}\cong \mathcal{O}(-n-1)$, obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.

For a fixed occupancy $n$, let $x_r>0$ be the per-flow rate on every active route, so feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. A proportionally fair allocation is a feasible allocation satisfying, for every other feasible $\tilde{x}$,The sum counts the proportional changes over individual flows. By the first-order condition for a concave function, it is exactly the allocation solvingThis is weighted logarithmic utility optimization, and its negative objective is a strictly convex function, which gives unique rates on active routes. Routes with $n_r=0$ have no per-flow rate to determine.

Index the four routes by $1=\{1,2\}$, $2=\{2,3\}$, $3=\{3,4\}$, $4=\{4,1\}$, and let ${\Lambda }_r=n_r{x}_r$ be each route's total service. WriteThe four capacity constraints say ${\Lambda }_1+{\Lambda }_2\le 1$, ${\Lambda }_2+{\Lambda }_3\le 1$, ${\Lambda }_3+{\Lambda }_4\le 1$, and ${\Lambda }_4+{\Lambda }_1\le 1$. Equivalently,For fixed maxima $u,v$, every active route in the first group optimally takes total service $u$, and every active route in the second takes $v$, because its logarithmic objective is increasing. Up to the constant $-{\sum }_r{n}_r\log n_r$, the objective therefore becomes $P\log u+Q\log v$ with $u+v\le 1$. Its maximizer is $u=P/N$, $v=Q/N$ when $N>0$, with the evident endpoint choices when one group is empty. Hence the proportionally fair allocation on a four-cycle isSet ${\Lambda }_r=0$ on inactive routes. If $N=0$, all service totals are zero. In particular, for $n_1>0$, $n_1{x}_1=P/N$, as required. Two disjoint active routes in the same group can each receive the same total service; service totals need not sum to one over the entire network.

With independent document arrivals given by Poisson processes of rates ${\rho }_r$ and independent unit-mean exponential distributions for document sizes, each of the $n_r$ active documents has completion rate $x_r$. The flow-level network model is consequently a continuous-time Markov chain withAt $n=0$ all downward rates are zero. The total departure rate is at most four, so the bounded total jump rate ensures no explosion.

The original PDF's displayed factor is a binomial coefficient, not the plain quotient produced by the TeX extraction. The candidate stationary distribution for this reversible four-cycle flow model isUse $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$. For $r\in \{1,3\}$ and $n_r>0$,so $\pi (n)q(n,n-e_r)=\pi (n-e_r){\rho }_r$. For $r\in \{2,4\}$ the corresponding ratio is ${\rho }_{r}N/Q$. Thus detailed balance for a continuous-time Markov chain holds for every adjacent pair, proving the displayed stationary distribution whenever $B<\infty$.

For completeness, its existence condition can be made explicit. Put $\alpha =\max ({\rho }_1,{\rho }_3)$ and $\beta =\max ({\rho }_2,{\rho }_4)$. At fixed $P,Q$,Therefore the sum over states with $N=m$ is at most $(m+1{)}^2(\alpha +\beta {)}^m$, using the binomial theorem. This is summable if $\alpha +\beta <1$. Conversely, restricting to the single maximizing route in each group leaves the series ${\sum }_{m\ge 0}(\alpha +\beta {)}^m$, which diverges if $\alpha +\beta \ge 1$. HenceThese are exactly the four strict resource-load inequalities. With positive arrival rates the Markov chain is irreducible, so the normalized stationary distribution is unique.

Use a positive-energy unitary representation with finite helicity states, excluding continuous-spin representations. In the specified frame the Pauli matrices giveFor every $A$ and every state $|\chi ⟩$,Positivity therefore gives $Q_2^A={\bar{Q}}_{\dot{2}A}=0$ on this massless supermultiplet. The equal-chirality anticommutator $\{Q_1^A,Q_2^B\}={ϵ}_{12}{Z}^{AB}$ then shows that every central charge in supersymmetry vanishes on this superalgebra representation.

Normalize the remaining supercharges as fermionic raising and lowering operators:They obey the canonical anticommutation relationsChoose a Clifford vacuum $|\lambda ⟩$ of helicity $\lambda \in \frac{1}{2}\mathbb{Z}$ annihilated by every $a_A$. Such a vector exists because the commuting fermion number operators $a_A^{†}{a}_A$ have eigenvalues $0$ and $1$, and lowering every occupied mode produces a nonzero empty state. In an irreducible representation this empty space has dimension one; multiple empty states produce a direct sum of copies. The states areThe canonical anticommutation relations make these states orthonormal for a normalized Clifford vacuum. In the helicity convention $[J_3,Q_1^A]=\frac{1}{2}{Q}_1^A$, its adjoint has $[J_3,a_A^{†}]=-\frac{1}{2}{a}_A^{†}$. Thus the Clifford vacuum is the highest-helicity state andReversing the rotation convention reverses the displayed ordering and leaves all counts unchanged. The level-$k$ states form the exterior power ${\Lambda }^k{\mathbb{C}}^{\mathcal{N}}$, which explains the binomial coefficient. Summing the binomial coefficients gives the helicity spectrum of a massless supermultiplet:Each application of a fermionic raising and lowering operator reverses fermion parity. The even and odd levels each contain $2^{\mathcal{N}-1}$ states, so a physical massless supermultiplet has $2^{\mathcal{N}-1}$ boson and $2^{\mathcal{N}-1}$ fermion states.

These formulas describe one irreducible representation of the massless Super-Poincaré algebra at fixed four-momentum. A reducible superalgebra representation can contain several such supermultiplets. Also, the CPT theorem sends helicity $h$ to $-h$ and conjugates internal charges. CPT completion of a supermultiplet may therefore require a second $2^{\mathcal{N}}$-state supermultiplet, with highest helicity $\mathcal{N}/2-\lambda$. A necessary condition for the original helicity spectrum of a massless supermultiplet to be self-conjugate is $\lambda =\mathcal{N}/4$; internal charges must also admit the conjugation in a realization compatible with the Spin-statistics theorem. Thus this helicity condition alone is not sufficient for CPT completion of a supermultiplet without doubling. In particular, the original irreducible count is not automatically the count after CPT completion of a supermultiplet.

For the conventional bounds on extended supersymmetry, suppose all massless states satisfy $|h|\le s_{\max }$. The helicity spectrum of a massless supermultiplet occupies an interval of width $\mathcal{N}/2$, whereas the available interval has width $2s_{\max }$. HenceAn ordinary four-dimensional renormalizable quantum field theory without gravity uses interacting massless fields of spin angular momentum at most one. Massless spin angular momentum $3/2$ requires supergravity, whose gravitational coupling has negative mass dimension and is a nonrenormalizable interaction by power counting in quantum field theory. Setting $s_{\max }=1$ gives $\mathcal{N}\le 4$ for conventional renormalizable theories. The bound is attained by four-dimensional N=4 super Yang-Mills theory: choosing $\lambda =1$ givesfor its supersymmetric vector multiplet.

Allowing gravity permits $s_{\max }=2$, so $\mathcal{N}\le 8$ for conventional theories including gravity. The four-dimensional N=8 supergravity supergravity multiplet attains the bound with $\lambda =2$:It has $256$ states, $128$ boson and $128$ fermion states. The conventional claim about a maximum “in general” assumes local interacting four-dimensional theories in Minkowski spacetime with a finite spectrum of fields and no massless spin angular momentum above two. It is not a theorem excluding arbitrary free higher-spin angular momentum constructions or all theories in other spacetime settings.

Finally, a chiral gauge spectrum means that the independent left-handed Weyl spinors occur in complex gauge group representations without obligatorily paired conjugate left-handed Weyl spinors. In four-dimensional N=1 supersymmetry, an irreducible matter supermultiplet with helicities $(0,-\frac{1}{2})$ can carry a complex gauge group representation $R$. Its CPT completion of a supermultiplet supplies the right-handed antiparticle in $\bar{R}$, as every left-handed Weyl spinor requires; it does not supply a second independent left-handed Weyl spinor in $\bar{R}$. Consequently a chiral gauge spectrum is possible.

For $\mathcal{N}=2$, the matter hypermultiplet has helicities $(\frac{1}{2},0,0,-\frac{1}{2})$ before internal charge conjugation. If it carries a complex gauge group representation $R$, its CPT completion of a supermultiplet adds the conjugate hypermultiplet. Equivalently, the resulting full hypermultiplet contains two $\mathcal{N}=1$ chiral superfields, one in $R$ and one in $\bar{R}$. These give a vectorlike gauge spectrum. Half-hypermultiplets can occur in pseudoreal representations, which also do not produce genuinely complex chiral gauge spectra. A supersymmetric vector multiplet has Adjoint representation, hence real gauge quantum numbers. Since every $\mathcal{N}\ge 2$ algebra contains an $\mathcal{N}=2$ subalgebra, the same obstruction applies to greater extended supersymmetry.

Therefore a conventional chiral gauge spectrum is possible only for $\mathcal{N}=0$ or $\mathcal{N}=1$. Here $\mathcal{N}=0$ means no supersymmetry; among theories with actual supersymmetry, only $\mathcal{N}=1$ qualifies. These are necessary structural conditions, not a guarantee that a chosen chiral gauge spectrum satisfies anomaly cancellation.

The hypothesis says exactly that $\mathcal{A}$ is a k-Sperner family with $k=2$: it contains no three members forming a strict inclusion chain. Every maximal chain in a Boolean lattice consequently contains at most two members of $\mathcal{A}$, so the same random-chain argument gives $\lambda (\mathcal{A})\le 2$. For a fixed amount of Lubell mass, cardinality is maximized by using the levels with the two largest binomial coefficients. The Erdős theorem on k-Sperner families therefore givesThe bound is attained by taking the union of two levels having those sizes.

For a uniform set family $\mathcal{A}\subseteq [n{]}^{(r)}$, the Local LYM inequality isTo prove it, count the incident pairs $(B,A)$ with $B\in \partial \mathcal{A}$, $A\in \mathcal{A}$, and $B\subset A$. Every $A$ contributes exactly $r$ pairs, whereas every $B$ belongs to at most $n-r+1$ members of $[n{]}^{(r)}$. Thereforewhich is the displayed inequality after using the binomial coefficient identity $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

The LYM inequality says that every antichain $\mathcal{A}\subseteq \mathcal{P}([n])$ satisfiesFor the proof from the local inequality, take the lowest occupied level below the middle and replace that level by its upper shadow. The result remains an antichain: any new containment involving a set in the upper shadow would already give a containment involving the member of $\mathcal{A}$ immediately below it. The dual form of the Local LYM inequality says that this replacement cannot decrease the Lubell mass. Repeating upward below the middle, and similarly replacing high levels by their lower shadows, eventually puts the whole family in one middle level. Its final Lubell mass is at most one, so the original mass is also at most one.

For the maximal chain in a Boolean lattice proof, choose a Uniformly random maximal chain in a Boolean lattice. An $h$-element set lies on it with probability $1/\left(\genfrac{}{}{0ex}{}{n}{h}\right)$. Because an antichain meets each chain at most once, the expected value of the number of its members on the chain is at most one. By linearity of expectation, that expected value is precisely the displayed sum.

The Sperner theorem follows because $\left(\genfrac{}{}{0ex}{}{n}{|A|}\right)\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ for every $A$. If equality holds in the resulting cardinality bound, every member lies on a largest level. For even $n$ this is the unique middle level. For odd $n$, the two middle levels have equal size; the regular connected inclusion graph between them and equality in Local LYM inequality force a chosen portion of the lower level to be either empty or the whole level. Hence the maximum antichains are exactly the complete middle level, with either middle level allowed when $n$ is odd.

By Bayes theorem, with respect to Lebesgue measure on $(0,1{)}^2$,The omitted constant includes the binomial coefficient and the normalizing constant of the Beta distribution; the uniform prior contributes only the support indicator.

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

Define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ as the number of $r$-element subsets of an $n$-element set, equivalently as $n!/[r!(n-r)!]$. Partitioning the $r$-element subsets of $\{1,\dots ,n+1\}$ according to whether they contain $n+1$ proves Pascal's identityThus the forward difference operator satisfies

For $0\le k\le n$, the binomial coefficient isWhen $1\le k\le n-1$,This is Pascal's identity.

The extension of the one-dimensional freely jointed chain isAs the integer $n_{+}$ runs from $0$ to $n$, the possible extensions are thereforeFor fixed $n_{+}$ and $n_{-}=n-n_{+}$, choosing which $n_{+}$ of the $n$ links point in the positive direction determines the microstate. The state degeneracy is consequently the binomial coefficient

The binomial theorem givesThe central binomial coefficient is the largest of these $2n+1$ nonnegative binomial coefficients, so

For the upper bound, fix a prime number $p\le 2n$. By Legendre formula,Each summand is either zero or one. Hence the P-adic valuation satisfiesIf $p\le \sqrt{2n}$, this impliesThere are at most $\sqrt{2n}$ such prime numbers. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the P-adic valuation is at most one. Multiplying the prime-power contributions therefore yields

We give the Agrawal–Biswas primality test, which has one-sided error. Small inputs and perfect powers can first be recognized deterministically. For every remaining integer $n$, put $d=\lceil 4{\log }_{2}n\rceil$ and choose a uniformly random monic polynomialUsing repeated squaring in the quotient ring $(\mathbb{Z}/n\mathbb{Z})[X]/(r)$, test the identityThis takes time polynomial in $\log n$ because every intermediate polynomial has degree below $d=O(\log n)$.

If $n$ is prime, the intermediate binomial coefficients are divisible by $n$, so the identity always holds. Now suppose that $n$ is composite and is not a prime power. Choose a prime divisor $p$ and write $n=p^{a}m$ with $p\nmid m$ and $m>1$. Over ${\mathbb{F}}_p$,because an intermediate coefficient equal to $m$ is nonzero modulo $p$. Henceis a nonzero polynomial of degree below $n$ over ${\mathbb{F}}_p$.

Reduction of random $r$ modulo $p$ is uniform among the $p^d$ monic degree-$d$ polynomials. The polynomial $F$ has at most $n/d$ distinct monic irreducible factors of degree $d$. On the other hand, the number $I_p(d)$ of monic irreducibles of degree $d$ obeys the standard lower boundWhenever $r\mathrm{mod}p$ is one of these irreducibles but does not divide $F$, the tested congruence fails. Thus one trial detects compositeness with probability at leastafter the finitely many small $n$ are handled directly. Repeating $2d$ times makes the probability of missing a composite less than $(1-1/(2d){)}^{2d}<1/2$, while a prime is never rejected. Therefore compositeness is in RP, and primality testing is in

For odd $p$ and $k\not\equiv 0(\mathrm{mod}p)$, take $E:y^2=x^3+kx$. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels the Legendre symbols. If $p\equiv 1(\mathrm{mod}4)$, the coefficient formula for trace of Frobenius modulo p givesThe binomial coefficient is nonzero because every factorial involved is shorter than $p$. Thus the Trace of Frobenius is not zero in the second case.

For a prime number $p$ and integer $r\ge 1$, every interior binomial coefficient in row $p^r$ is divisible by $p$. The polynomial identity $(1+x{)}^p\equiv 1+x^p(\mathrm{mod}p)$ iterates under the Frobenius endomorphism to $(1+x{)}^{p^r}\equiv 1+x^{p^r}(\mathrm{mod}p)$. This concerns divisibility by $p$, not necessarily by $p^r$.

For the proportionally fair allocation on a linear flow network, independent Poisson flow arrivals at rates ${\nu }_r$ and exponential document sizes of rates ${\mu }_r$ give a continuous-time Markov chain with departures ${\mu }_r{n}_r{x}_r(n)$. Put ${\rho }_r={\nu }_r/{\mu }_r$. Under ${\rho }_0+{\rho }_i<1$ for each local route, the normalizing constant is $Z=(1-{\rho }_0{)}^{I-1}/{\prod }_i(1-{\rho }_0-{\rho }_i)$. The binomial coefficient weight has adjacent-state ratios equal to the aggregate service rates, proving detailed balance. For more general network topologies, proportional fairness need not give this reversible product form.

The uniform layer consists of the characteristic vectors of sets of size $r$ in $[n]$. It has size $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, a binomial coefficient. Restricting polynomials to a uniform layer introduces identities absent on the entire Boolean lattice, since the coordinate sum is fixed. In particular, homogenisation on a uniform layer spans all restricted multilinear polynomials of degree at most $s\le r$ using just the degree-$s$ monomials.