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In the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, choose $e$ coprime to $\phi (N)=(p-1)(q-1)$, and choose $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. Publish $(N,e)$ and keep the factorization and $d$ private. Encrypt a properly encoded message $M$ as $C=M^e\mathrm{mod}N$ and decrypt as $M=C^d\mathrm{mod}N$. The Chinese remainder theorem proves correctness even for nonunits: modulo either prime, the message is either zero or satisfies Fermat's little theorem.

The public exponent $65537=2^{16}+1$ has few nonzero binary digits, so exponentiation needs only sixteen squarings and one extra multiplication. It is an efficient conventional choice, subject to the required coprimality and secure randomized message encoding. The same numerical choice for the private exponent would speed decryption, but for large moduli with balanced prime factors it is disastrously small: the continued-fraction attack on a small RSA private exponent recovers $d$ when, for example, $d<\frac{1}{3}{N}^{1/4}$ and $q<p<2q$. A small public exponent and a small private exponent have very different security implications.

To factor $N$ from $e,d$, write $ed-1=2^{s}t$ with $t$ odd. Choose random $a$ modulo $N$. If $\gcd (a,N)$ is nontrivial, factoring is already done. Otherwise compute $b=a^t$ and repeatedly square; the final value is one. If a value $x$ immediately preceding the first one is neither $1$ nor $-1$ modulo $N$, then $x^2\equiv 1(\mathrm{mod}N)$ andIndeed, the signs of $x$ modulo $p$ and $q$ differ. Restart if the chain begins at one or reaches $-1$ first. The Chinese remainder theorem makes the two prime components independent for a uniformly chosen unit. Their two-primary orders differ with probability at least one half: if the two exponents of two in $p-1,q-1$ are equal, the probability of equal orders is at most one half, and if unequal it is smaller. Since odd $t$ preserves these orders, this gives a randomized factoring procedure with bounded expected repetitions.

A bit commitment must be hiding before opening and binding after it. A concrete RSA construction uses a valid public RSA permutation whose inverse is unavailable to the receiver. Choose a random unit $x$ and a random bit vector $r$, and set $h(x,r)$ to their binary inner product modulo two. Send $y=x^e\mathrm{mod}N$, $r$, and $c=b\oplus h(x,r)$. To open, reveal $x$; the receiver checks $x^e=y$ and recovers $b=c\oplus h(x,r)$. Injectivity makes the opening bit unique. The hard-core predicate theorem applied to $(x,r)\mapsto (x^e,r)$ gives computational hiding under the RSA inversion assumption. The setup must ensure a valid modulus and exponent without giving the receiver the private exponent; otherwise hiding fails. Merely encrypting $0$ or $1$ would not hide a bit, since both ciphertexts could be computed publicly.

The Prime number theorem states that $\pi (x)\sim x/\log x$, where $\pi (x)$ counts primes not exceeding $x$. Dirichlet theorem on primes in arithmetic progressions states that if $\gcd (a,m)=1$, then the progression $a+jm$ contains infinitely many primes.

If $x^2\equiv -1(\mathrm{mod}p)$ for an odd prime $p$, the multiplicative order of $x$ modulo $p$ is four. Lagrange's theorem therefore gives $4\mid p-1$. Conversely, Wilson's theorem gives $(p-1)!\equiv -1(\mathrm{mod}p)$. Pairing the factors $j$ and $p-j$ yieldsFor $p\equiv 1(\mathrm{mod}4)$, the sign is positive, so the factorial supplies a square root of $-1$. Thus $-1$ is a quadratic residue exactly when $p\equiv 1(\mathrm{mod}4)$.

Write $P={\prod }_j{p}_j$. The odd integer $4P-1$ is $3$ modulo $4$. If all of its prime factors were $1$ modulo $4$, their product, with multiplicities, would be $1$ modulo $4$, so at least one factor is $3$ modulo $4$. Every prime factor $q$ of $4P^2+1$ is odd and satisfies $(2P{)}^2\equiv -1(\mathrm{mod}q)$; the criterion just proved gives $q\equiv 1(\mathrm{mod}4)$. Neither integer is divisible by any $p_j$, since their residues modulo $p_j$ are respectively $-1$ and $1$. Starting with any finite list thus produces a new prime in each residue class. Both classes contain infinitely many primes, independently of the general Dirichlet theorem on primes in arithmetic progressions.

The kinetic energy is $m({\dot{\eta }}_1^2+{\dot{\eta }}_2^2)/2$. The Euler–Lagrange equations giveFor a normal mode $\eta =A{e}^{i\omega t}$, the stiffness matrix isSolving this quadratic yieldsBoth are positive for positive spring constants. A corresponding amplitude ratio is $A_2/A_1=(j+k-m{\omega }_{\pm }^2)/k$; the lower-frequency mode has the masses moving in the same direction and the higher-frequency mode in opposite directions.

Take a Hankel contour $C_{\eta }$ around the positive real axis, going from infinity on its lower side toward a circle of radius $0<\eta <1$, clockwise around that circle, then out along the upper side. Use $0\le \arg z\le 2\pi$ and defineThis is an entire function of $t$: the small circle stays away from zero and the pole at $-1$, while the exponentially decaying rays converge uniformly for $t$ in compact subsets. Contour deformation shows independence of $\eta$. For $\Re t>-1$, let $\eta \to 0$. The upper ray contributes $F(t)$, the lower ray contributes $-e^{2\pi it}F(t)$, and the small-circle contribution vanishes. Thus the analytic continuation isThe only apparent singularities are at integers. At the nonnegative integers they are removable because the original integral is holomorphic throughout $\Re t>-1$. Hence singularities can occur only at $-1,-2,\dots$.

In fact all of those points are simple poles. At $t=-n-1$ the two rays cancel and the clockwise circle gives $H(-n-1)=-2\pi i{c}_n$, whereSince the denominator has derivative $-2\pi i$ at each integer, the residue at $-n-1$ is $c_n$. The parameter has been called $t$ consistently; the final use of $z$ for this parameter in the printed question is harmless notation.

A strict Lyapunov function is a continuously differentiable function $V$ with $V(0)=0$, $V(x)>0$ off the origin, and $\nabla V\cdot f<0$ off the origin in the domain. Its compact sublevel sets inside the domain are forward invariant and force convergence to the equilibrium. The domain of stability, or basin of attraction, consists of initial points whose forward solutions exist for all positive time and converge to the fixed point.

For $V=(x^2+y^2)/2$, direct differentiation and factorization giveIf $R^2=x^2+y^2<1/2$, then $x^2+4y^2\le 4R^2<2$, so $\dot{V}<0$ off zero. For any initial point in that disc, choose a slightly larger closed disc still inside it. The vector field points inward on its boundary; the trajectory is bounded, exists globally, and the strict Lyapunov decrease forces its limit to be zero. Thus the origin is asymptotically stable and the whole open disc lies in its basin.

If $R^2>2$, then $x^2+4y^2\ge R^2>2$, so $\dot{V}>0$. A solution starting there cannot cross inward through $R^2=2$ and cannot approach zero. Therefore

The prey have per-capita growth ${\mu }_1$, loss through encounters with predators at rate ${\alpha }_{1}v$, and density-dependent competition $\delta u$. Predators suffer per-capita mortality ${\mu }_2$, gain from consuming prey at rate ${\alpha }_{2}u$, and undergo constant harvesting or removal $ϵ$. Constant harvesting makes the equations biologically valid only while the predator population remains nonnegative.

At an equilibrium with positive populations,Writing $A={\alpha }_2{\mu }_1-\delta {\mu }_2>0$ givesFor $0<ϵ<A^2/(4{\alpha }_1{\alpha }_2\delta )$, both roots are positive and give $u>{\mu }_2/{\alpha }_2>0$ via the second equilibrium equation. Thus there are two positive equilibria.

Using the equilibrium identities in the Jacobian givesThe smaller root satisfies $v_{-}{v}_{+}=\delta ϵ/({\alpha }_1{\alpha }_2)$ and $v_{-}<v_{+}$, so $v_{-}^2<\delta ϵ/({\alpha }_1{\alpha }_2)$. Hence its determinant is negative: it is always a saddle and unstable. For small $\delta ,ϵ$, this is the equilibrium with $u\sim {\mu }_1/\delta$ and $v\sim \delta ϵ/({\alpha }_2{\mu }_1)$, as required. The determinant argument is stronger than merely taking the small-parameter limit.

Independence and the Poisson probabilities give the log-likelihoodIts score and curvature areIf some $x_i\ne 0$, this is strictly concave. A finite maximum is therefore the unique zero of the score when such a zero exists. Newton iteration givesStart from a finite value, compute the score and curvature, and iterate until both the score and step are small. A line search that shortens a step until the likelihood increases improves robustness. Strict concavity identifies any converged stationary point as the global maximum. A finite root need not exist, for example when all observations are zero and all $x_i>0$, in which case the likelihood supremum is at $\beta \to -\infty$. If all $x_i=0$, the parameter is unidentifiable.

A decipherable code is one whose extension to finite strings by concatenation is injective: an encoded string has a unique sequence of source letters. In particular its codewords are distinct and nonempty. Put $K={\sum }_i{a}^{-s_i}$ and $s_{\max }={\max }_i{s}_i$. For strings of exactly $n$ source letters, let $N_j$ count the encoded strings of length $j$. Unique decipherability gives $N_j\le a^j$, since distinct source strings yield distinct encoded strings. Expanding $K^n$ givesTaking $n$th roots and letting $n\to \infty$ proves the McMillan inequality

Add $00,01,10$ to the four specified binary words. None of the resulting seven words is a proper suffix of another: the longer words end in $11$, and their shorter terminal strings never equal another of the specified longer words. Thus this is a suffix code. Reading from the right determines the last codeword uniquely; deleting it and repeating proves unique decipherability. Its length sum is $3/4+1/8+1/16+2/32=1$, in agreement with the bound.

For a nonidentity element of ${\mathrm{PSL}}_2(\mathbb{R})$ represented by a determinant-one matrix, the classification is elliptic, parabolic or hyperbolic according as the squared trace is less than, equal to or greater than four. This follows either from the eigenvalues or from the discriminant of its fixed-point equation on the boundary of the upper half-plane.

The first matrix has determinant one and trace two, and is not scalar. It is therefore parabolic. Its fixed-point equation isThus its unique fixed point is $z=-2$ on the real boundary, of multiplicity two; infinity is not fixed and there is no fixed point in the upper half-plane. The second matrix has determinant one and trace $-4$, so it is hyperbolic. Its two boundary fixed points are $(-1\pm \sqrt{3})/2$, consistently with this classification.

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.

The initial Fourier transform is $\hat{\zeta }(k,0)=2b{\zeta }_0/(1+b^2{k}^2)$. Initial rest gives zero time derivative, so each wave mode evolves by $\cos (\sqrt{g|k|}t)$. Fourier inversion therefore givesOn $x=Vt$, the product of cosines produces phases ${\phi }_{\pm }(k)=Vk\pm \sqrt{gk}$ with prefactor $b{\zeta }_0/\pi$. Only the minus phase has a stationary point:Expanding quadratically there and evaluating the Gaussian oscillatory integral gives the stationary-phase factor $\sqrt{2\pi /(t{\phi }_{-}^{\prime \prime }(k_{\ast }))}$ and phase shift $+\pi /4$. The other phase has no stationary point. Using $q=\sqrt{k}$ makes both endpoint phases smooth with nonzero endpoint derivative and amplitude vanishing at zero, so endpoint contributions are smaller. ThusThe selected wavenumber has group velocity $V$, explaining why it determines the long-time signal on this ray.

In the lubrication regime the thin region has longitudinal scale $\ell \sim \sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell /b\sim \mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.

Use a frame translating with the cylinder. The wall moves with velocity $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, soContinuity makes $q$ constant. Equal ambient pressures at both infinities give $\int p_{x}dx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. ThereforeThe vertical velocity, with the wall condition $w(0)=0$, follows explicitly from incompressibility:It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.

The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}s$, its integral isForce balance equates this wall force to the required cylinder force at leading order; the pressure and viscous stresses on remote boundaries make no leading contribution. This is the cylinder translating parallel to a wall in a thin gap result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.