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Let $N=pq$ with primes $p\equiv 1(\mathrm{mod}12)$ and $q\equiv 5(\mathrm{mod}12)$. quadratic reciprocity makes three a square modulo $p$; choose $r^2=3$ and impose $a=-3+2r(\mathrm{mod}p)$, $a=1(\mathrm{mod}q)$ by the Chinese remainder theorem. The curve has good reduction at both primes. Indeed its discriminant is $-16a^2(4a+27)$; the first condition could make a factor vanish only at $p=3$ or 11, while the second gives $a(4a+27)=31$ and the given residue class excludes $q=31$.

Doubling $P$ gives $x(2P)-1=(a^2+6a-3)/4$. This vanishes modulo $p$ and equals one modulo $q$. Modulo $p$ one also has $y(2P)=-1$, so $2P=-P$ and $3P$ is the identity. Modulo $q$ the addition is finite. Thus computing $3P$ modulo $N$ encounters a denominator whose greatest common divisor with $N$ is exactly $p$, revealing the factor. The numerator difference is minus two modulo $p$, so the failed inverse cannot be cancelled. This is a construction of successful elliptic curve method data from prescribed local group behavior.

Let $p\ge 3$ be a prime number dividing $n^2+3$, and set $\omega =(-1+i\sqrt{3})/2$. The case $p=3$ is $N(1-\omega )=3$. For $p>3$, useThe Eisenstein integer $p$ divides this product, but divides neither factor: divisibility by the rational integer $p$ in the basis $1,\omega$ requires both integer coefficients to be divisible by $p$, whereas the second coefficients are $2$ and $-2$. Thus $p$ is not a prime element of $\mathbb{Z}[\omega ]$.

The Euclidean norm on the Eisenstein integers makes this ring a principal ideal domain. Since irreducible elements of a principal ideal domain are prime, $p$ admits a factorization $p=uv$ into two nonunits. The positive integral Eisenstein norm is multiplicative, and its value is one exactly on units: when $N(u)=1$, its complex conjugate is an inverse. Thus $N(u),N(v)>1$ and $N(u)N(v)=p^2$, forcing both norms to be $p$. Writing $u=a+b\omega$ proves the displayed representation.

On an open set where $u>0$, put $v=\sqrt{u}$. The chain rule gives $\nabla u=2v\nabla v$, so the quadratic eikonal equation is equivalent to $|\nabla v|=1$. For constant positive data $v=v_0$ on a smooth initial curve, the tangential derivative vanishes and the two allowed normal derivatives are ${\partial }_{n}v=\pm 1$. Locally the two solutions are $v=v_0\pm d$, where $d$ is oriented distance along the normal rays; hence $u=(v_0\pm d{)}^2$ wherever the chosen $v$ stays positive and the normal-coordinate map remains invertible.

There are two separate obstructions to using this initial-data chart. A characteristic caustic occurs when normal rays focus and the projection Jacobian determinant vanishes. Alternatively $v$ can reach zero while the geometric rays remain separated: the square-root transformation then ceases to apply and the original Charpit's equations have a degenerate zero-gradient state. A continuation of $u$ beyond either obstruction needs separate justification; squaring an algebraic expression alone does not preserve a prescribed positive square-root branch.

This map is a diffeomorphism from ${\mathbb{R}}^2$ onto $V=\{(s,t):s>0,t>0,st>1\}$. Writing $q=e^x$, the inverse problem becomes $(s-q)(t-q)=1$ with $0<q<\min (s,t)$. The left side decreases strictly from $st$ to zero, proving existence and uniqueness exactly on $V$. Its Jacobian determinant is $-e^x(e^y+e^{-y})\ne 0$, so the inverse function theorem supplies a differentiable inverse. Along $(x,y)=(c,c)$, applying the inverse Jacobian to $(1/2,1/2)$ gives $(e^{-c}/2,0)$. Establishing the actual image matters before using a global inverse, even when a local inverse already exists at the point of interest.

Convert a curried function into one taking a tuple: uncurry f (x,y) = f x y. Conversely, curry f x y = f (x,y). The two conversions undo each other on total arguments. Uncurrying lets map apply a curried binary operation to each tuple in a zipped list.

In functional programming, currying represents a function of several arguments as successive one-argument functions. In Haskell, arrows associate to the right: a -> b -> c means a -> (b -> c). Supplying an argument of type a therefore returns a function of type b -> c, while a function of type (a,b) -> c expects a tuple. Uncurrying converts the first representation to the second.

For a nonconstant entire function, its derivative is nonzero somewhere and therefore bounded away from zero on a small disc. Recurrence of planar Brownian motion and the Strong Markov property imply infinite total occupation time in that disc: repeat entrances into a smaller concentric disc after exits from the larger one. On each entrance there is a uniformly positive chance of remaining inside the larger disc for a fixed positive duration. Iterating the conditional failure bound gives infinitely many such sojourns almost surely.

The displayed integral is bounded below by a positive constant times this occupation time, so it diverges. The same conclusion follows by comparing with the divergence of a positive planar Brownian occupation integral. At every finite time the clock is finite, because the Brownian path has compact range and $f^{\prime }$ is continuous. Thus the conformal Brownian clock covers every nonnegative image time; conformal invariance of planar Brownian motion then yields a Brownian image with no missing finite or infinite lifetime.

If $f$ is twice continuously differentiable on $[a,b]$, take a uniform bound $|f^{\prime \prime }|\le M$. On one panel $[u,v]$, two integrations by parts give the signed quadrature errorThe nonnegative kernel has integral $(v-u{)}^3/6$, so the absolute error is at most $M(v-u{)}^3/12$. Summing $n$ equal panels proves the displayed bound. The maximum exists on the closed interval; a supremum or any known uniform bound also suffices. The estimate certifies a sufficient panel count, which need not be the least count achieving a particular actual error.

On a uniform partition of $[a,b]$ with width $h$, the right Riemann sum and the composite trapezoidal rule have equal weights at every interior mesh point. They differ only by adding half a final-endpoint weight and removing half an initial-endpoint weight. This proves the displayed identity. For continuous $f$, both converge to its definite integral; their finite endpoint correction need not be confused with the smaller trapezoidal discretization error for a smooth integrand.

On a partition of an interval symmetric about zero, the composite trapezoidal rule gives opposite quadrature nodes equal quadrature weights. Samples of an odd function cancel pairwise, and a zero quadrature node contributes zero. For a continuous odd function, its exact integral over $[-L,L]$ is also zero. Adding any affine function preserves exactness by linearity of the integral and of the quadrature rule. For example, $1+\sin x$ is nonpolynomial but both the exact integral and every uniform-grid composite trapezoidal rule value on $[-1,1]$ are two.

A basis of a matrix's column space. Row reduction identifies which original columns to select: take the original columns at the pivot columns of the reduced matrix. Row operations preserve column dependence relations, but generally change the column vectors and their span. Choosing columns of the reduced matrix instead therefore need not give the original column space.

In $x,y>0$, the change of variables in the title reducesto the one-dimensional wave equation $U_{ss}-U_{rr}=0$. Indeed $u_{xx}=U_{rr}/(4x)-U_r/(4x^{3/2})$ and $u_{yy}=y{U}_{ss}/4+U_s/(4\sqrt{y})$; the first-derivative terms cancel exactly against the right-hand side. The independent characteristic coordinates $\xi =r+s$, $\eta =r-s$ then give $V_{\xi \eta }=0$ and $u=g(\xi )+h(\eta )$. Two characteristic families alone generally reduce only the principal part; the cancellation of lower-order terms is the extra feature that gives this simple exact solution.

Every object $M$ of Category O has a finite composition series of a module. Choose weight-vector generators and let $T$ be the sum of their finite-dimensional weight spaces. The centre preserves $T$; the kernel of its action on $T$ has finite codimension and, by centrality, annihilates all generators and hence all of $M$. Thus only finitely many central characters can appear in subquotients. The Harish-Chandra isomorphism makes each such character correspond to one finite Weyl orbit of shifted highest weights. Let $S$ be the resulting finite set of possible highest weights.

Every nonzero subquotient has a highest-weight vector from its bounded-above support, and a common central eigenvector can be chosen in the finite-dimensional highest-weight space. Its weight belongs to $S$. The integer $d(N)={\sum }_{\nu \in S}\dim N_{\nu }$ is therefore finite, additive on short exact sequences and positive for every nonzero subquotient $N$. Each strict refinement of a submodule chain consumes at least one unit of $d(M)$, bounding its length and proving the claim.

For knot spacing $h=1/2$ on $[0,1]$, the five translates indexed $-3,-2,-1,0,1$ span the $C^2$ piecewise cubic space. With normalized knot values $1/6,2/3,1/6$, nodal interpolation gives $(c_{-3}+4c_{-2}+c_{-1})/6=u(0)$ and the two shifted equations at $h,2h$. Left-end Hermite data give $(c_{-1}-c_{-3})/(2h)=u^{\prime }(0)$ and $(c_{-3}-2c_{-2}+c_{-1})/h^2=u^{\prime \prime }(0)$. These five equations uniquely determine the coefficients. Derivative data need existence or separate prescription; continuity of $u$ alone does not supply them.

For the centered Cardinal cubic B-spline $\beta$ with knot values $1/6,4/6,1/6$, write $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$ on $[0,mh]$. interpolation gives $c_{i-1}+4c_i+c_{i+1}=6u(ih)$ for $i=0,\dots ,m$. Since ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, the natural endpoint conditions $s^{\prime \prime }(0)=s^{\prime \prime }(mh)=0$ give the displayed extra two rows. These supply the two equations missing from interpolation alone.

A continuous nondecreasing function $\Phi :[0,\infty )\to [0,\infty )$ with $\Phi (0)=0$ has moderate growth if the displayed inequality holds for some $a>1$ and finite $K$, uniformly in $x\ge 0$. Iteration gives a bound for any fixed positive dilation, and at most polynomial growth at infinity. The powers $x^p$, $p>0$, satisfy the condition. It allows a good-lambda tail comparison to be integrated into a moment comparison.

For Brownian motion started at zero, condition at time ${\lambda }^2$. If the stopping time exceeds ${\beta }^2{\lambda }^2$ while the running absolute maximum stays below $\delta \lambda$, the path must remain in a fixed interval for another $({\beta }^2-1){\lambda }^2$ time units. Its expected exit time from this interval is at most ${\delta }^2{\lambda }^2$. Markov's inequality yields the displayed bound. Integrating it against $d\Phi$ for a function of moderate growth, then absorbing the small tail term, proves $\mathbb{E}\Phi (\sqrt{\tau })\le C_{\Phi }\mathbb{E}\Phi (B_{\tau }^{\ast })$.

Real and imaginary parts of an analytic Brownian image, after subtracting their starting values, have equal quadratic variation. The Burkholder-Davis-Gundy inequalities compare their centered maximal moments with constants depending only on $p$. A boundary-point normalization $P(0)=0$ does not remove the starting imaginary value. For $P(z)=iz$ on a wedge of opening $\alpha \pi <\pi /2$, $\sup |\mathrm{Im}z|\le \tan (\alpha \pi )\sup |\mathrm{Re}z|$. If $\alpha p<1$ both relevant moments are finite, so an uncentered comparison requires a constant at least ${\cot }^p(\alpha \pi )$; no bound depending only on $p$ can hold over all such wedges. For fixed $\alpha >0$, polynomials with finite real-part maximal moment have degree $d<1/(\alpha p)$ by the wedge radial-maximum tail. They form a finite-dimensional vector space, and equivalence of norms in finite dimensions gives a valid constant depending on $p,\alpha$.

On an infinite set $D$, take its finite and cofinite subsets with ordinary union, intersection and complement. Singletons are its atoms. Its Stone space consists of the isolated points of $D$ together with one nonisolated point whose ultrafilter contains all cofinite sets. For countably infinite $D$ this is a countable Boolean algebra with countably many atoms, not the uncountable full power set. It describes two-vector definability over a countably infinite field: dependent directions are isolated atoms and the independent-pair type is the additional nonisolated point.

An atom is a minimal nonzero Boolean-algebra element. In a Lindenbaum algebra, its nonzero formula class isolates a complete type. A finite Boolean algebra is the power set of its finite atom set. An infinite atomic Boolean algebra need not contain every union of atoms: the finite-cofinite algebra is a simple counterexample.