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The interface is a material surface, so its defining function $F=z-\zeta (x,y,t)$ obeys $DF/Dt=0$ on each side. Since ${\mathbf{u}}_i=\nabla {\Phi }_i$, the exact kinematic boundary condition isSubstitute ${\Phi }_i=U_{i}x+{\phi }_i$. Products such as ${\phi }_{ix}{\zeta }_x$ are second order, as is the correction from evaluating ${\phi }_{iz}$ at $z=\zeta$ instead of zero. Linearization therefore gives

For the charge-generated electric field, use the electric potentialThe vector potential is time independent, so its usual additional contribution $-{\partial }_t\mathbf{A}$ is zero. The continuity equation now givesThus the second term in part (iii) is exactly the displacement current contribution ${\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$, andThis is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.

Use the identity $\nabla \times \mathbf{B}=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$. The Laplacian identity for the Coulomb kernel gives, component by component,Also $\nabla (s^{-1})=-{\nabla }^{\prime }(s^{-1})$, so integration by parts yieldsHere the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives

Let $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$ and introduce the magnetic vector potential $\mathbf{A}(\mathbf{r})={\mu }_0(4\pi {)}^{-1}\int \mathbf{J}({\mathbf{r}}^{\prime })/sd{V}^{\prime }$. Differentiation is with respect to $\mathbf{r}$, so $\mathbf{J}({\mathbf{r}}^{\prime })$ is constant under it. HenceTaking the curl through the integral reproduces the Biot-Savart law, proving $\boxed{\mathbf{B}=\nabla \times \mathbf{A}}$ and the requested representation.

Charge conservation gives the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. Since $\mathbf{J}$ is independent of time, so is ${\partial }_t\rho =-\nabla \cdot \mathbf{J}$. Integrating at each fixed position gives

Use the same indented upper-semicircle contour and branch of the complex logarithm, now for $F(z)=z^{1/2}\mathrm{Log}z/(1+z^2)$. Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. On the negative axis, $z^{1/2}=i|z{|}^{1/2}$ and $\mathrm{Log}z=\log |z|+i\pi$. Consequently the two straight contributions tend toThe outer arc is $O(R^{-1/2}\log R)$ and the small arc is $O({\epsilon }^{3/2}(|\log \epsilon |+1))$, so neither contributes in the limit. At the enclosed pole $i$,The residue theorem gives $(1+i)J-\pi I(1/2)=i{\pi }^2{e}^{i\pi /4}/2$. Part (a) gives $I(1/2)=\pi /\sqrt{2}$. Substituting and using $e^{i\pi /4}=(1+i)/\sqrt{2}$, the right side after moving the $I$ term is ${\pi }^2(1+i)/(2\sqrt{2})$. Hence

Put $a=1/n$, so $0<a<1$, and write $I(a)={\int }_0^{\infty }{x}^a/(1+x^2)dx$. Choose the branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, whose cut is the downward imaginary axis. Define $z^a=\exp (a\mathrm{Log}z)$ and integrate $z^a/(1+z^2)$ along the positively oriented upper semicircle, indented above zero. The straight pieces run from $-R$ to $-\epsilon$ and from $\epsilon$ to $R$; the small arc runs clockwise from argument $\pi$ to zero.

On the negative real axis $z^a=|z{|}^a{e}^{i\pi a}$, so the two straight pieces tend to $(1+e^{i\pi a})I(a)$. The outer arc is $O(R^{a-1})$ and the inner arc is $O({\epsilon }^{a+1})$, both tending to zero. The only enclosed pole is $i$, with residue $e^{i\pi a/2}/(2i)$. The residue theorem therefore givesUsing $1+e^{i\pi a}=2e^{i\pi a/2}\cos (\pi a/2)$ yieldsThe same bounds show convergence of the real integral at both zero and infinity.

A Hausdorff, noncompact topology. The condition says precisely that each point of $U$ lies in a residue class $n+k\mathbb{Z}$ contained in $U$. These classes form a basis: the intersection of two classes containing a common point contains the class through that point with modulus equal to the least common multiple of their moduli. Thus arbitrary unions obey the condition, and so do finite intersections. Also $\varnothing$ and $\mathbb{Z}$ obey it. This is the profinite topology on $\mathbb{Z}$.

If $a\ne b$, choose $k>|a-b|$. Then $a+k\mathbb{Z}$ and $b+k\mathbb{Z}$ are disjoint open neighbourhoods, proving the Hausdorff property. To test compactness, defineEvery $F_n$ is nonempty and both open and closed: its complement is a finite union of other residue classes. Since $a_{n+1}-a_n=4^n$, $F_{n+1}\subseteq F_n$. If $x$ belonged to their intersection, then $3x+1$ would be divisible by $4^n$ for every $n$, forcing $3x+1=0$, impossible for an integer. Therefore $U_n=\mathbb{Z}\setminus F_n$ is an increasing open cover of $\mathbb{Z}$. Any finite subcollection has union $U_N$ for its largest index and misses the nonempty $F_N$. This proves that the profinite topology on the integers is not compact.

Not a topology. For every integer $k\ge 2$, the set $U_k=\mathbb{Z}\setminus k\mathbb{Z}$ is periodic with period $k$, so belongs to the family. Their union is $\mathbb{Z}\setminus \{0\}$: for a nonzero $n$, choose $k>|n|$, which cannot divide $n$. But this union has no positive integer period. If it had period $k$, membership of $-k$ would force membership of zero. Again arbitrary unions fail the topology axiom.

A Hausdorff compact topology. The empty set omits zero and $\mathbb{Z}$ has finite complement. An arbitrary union omitting zero belongs to the family. If a union contains zero, one of its constituent sets contains zero and is cofinite, so the union is also cofinite. A finite intersection either omits zero or is an intersection of finitely many cofinite sets and is therefore cofinite. This proves all topology axioms.

For distinct nonzero integers $a,b$, the disjoint open singletons $\{a\},\{b\}$ separate them. To separate $0$ and $n\ne 0$, use the disjoint open sets $\mathbb{Z}\setminus \{n\}$ and $\{n\}$. Thus this is a Hausdorff space. Every open cover has a member containing zero, which leaves only finitely many integers uncovered; choose one member for each of those points. This gives a finite subcover and proves compactness. The space is the one-point compactification of the discrete space $\mathbb{Z}\setminus \{0\}$, with zero as the added point.

Not a topology. Each singleton $\{n\}$ with $n>0$ is finite and hence belongs to the proposed family, but their union $\{1,2,3,\dots \}$ is infinite and excludes zero, so does not belong. The arbitrary-union axiom fails.

For each involution $t\in G$, its conjugacy class has size fifteen, so its centralizer $C_G(t)$ has order four. A group of order four is a cyclic group or a Klein four-group; the cyclic possibility would contain an element of order four, excluded by the geometric enumeration. Hence each $C_G(t)$ is a Klein four-group containing three involutions.

If $u$ is any of the three involutions in $C_G(t)$, the latter group is an abelian group and hence is contained in $C_G(u)$. Both centralizers have order four, so they are equal. It follows that the fifteen involutions partition into five triples, with five distinct centralizers. Conjugation by $G$ permutes these five subgroups and gives a group homomorphism $\theta :G\to S_5$. This action is nontrivial: otherwise each of these order-four subgroups would be a normal subgroup, contrary to simplicity. The kernel of $\theta$ is normal; since $G$ is simple and the action nontrivial, its kernel is trivial. Thus $\theta$ is injective.

The sign of $\theta (g)$ defines a homomorphism $G\to \{\pm 1\}$. If it were nontrivial, simplicity would make it injective, impossible for a group of order sixty mapping into one of order two. Hence every image permutation is even, and $\theta (G)\subseteq A_5$. Both groups have order sixty, so they are equal. This constructs the requested group isomorphism:

Write $G=\mathrm{Rot}(D)$ for the rotational symmetry group of a regular dodecahedron. Centre the solid at the origin. A rotation in three dimensions is determined by the image of a chosen face and one of its vertices, so there are at most $12\cdot 5=60$ rotations. All these choices occur. The outward unit normal to a face and the direction from its centre to the chosen vertex determine an oriented orthonormal frame. The proper rotation mapping one such frame to another maps the whole regular pentagonal face. At each edge, the common dihedral angle and convexity then determine the adjacent face uniquely; propagating across the connected face-adjacency graph shows that the rotation maps the entire solid to itself. Hence $|G|=60$, and the action is transitive on faces, vertices and edges.

Its rotations can be counted geometrically. There are six axes joining opposite face centres, each allowing four nonidentity rotations of order five; ten axes joining opposite vertices, each allowing two of order three; and fifteen axes joining opposite edge midpoints, each allowing a half-turn. The orbit-stabilizer theorem also verifies these possibilities: vertex stabilizers have order $60/20=3$ and edge stabilizers have order $60/30=2$, and their nonidentity elements rotate about the corresponding vertex or edge-midpoint axis. Together with the identity this accounts for all $1+24+20+15=60$ elements. In particular there are no elements of order four.

We now determine the conjugacy classes. The group is transitive on edges, so all fifteen edge half-turns are conjugate. For a rotation whose angle is neither zero nor $\pi$, a commuting rotation must preserve its oriented axis: conjugation carries the axis and its rotation angle to their images, and reversing the axis would reverse a non-half-turn angle. Thus its centralizer consists of rotations about the same axis. For a face rotation this centralizer has order five, and for a vertex rotation it has order three. The orbit-stabilizer theorem for conjugation therefore gives class sizes $60/5=12$ and $60/3=20$. The twenty vertex rotations form one class, and the twenty-four face rotations form two classes. The class sizes are $1,15,20,12,12$.

A normal subgroup is a union of these classes and contains the identity. Its possible orders areBy Lagrange's theorem its order must divide $60$. Only $1$ and $60$ in this list do so. Therefore $\boxed{G\text{ is simple of order }60}$.

The alternating group $A_4$ containsThe product of any two different nonidentity elements is the third, and each squares to the identity, so $V$ is a Klein four-group. Conjugation by a permutation replaces the labels in each cycle, preserving the property of being a product of two disjoint transpositions. Thus $V$ is a normal subgroup of $A_4$. Its order is $4$, strictly between $1$ and $|A_4|=12$. $A_4$ is not simple.

Suppose $x^2\equiv -1(\mathrm{mod}p)$. Then $x\ne 0$ modulo the prime number $p$, so Fermat's little theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. But $p=4m+3$ makes $(p-1)/2=2m+1$ odd, and thereforeThis contradicts $1\not\equiv -1(\mathrm{mod}p)$, since $p$ is odd. Thus $-1$ is not a quadratic residue modulo $p$. Apply part (a) to the finite field $F={\mathbb{F}}_p$: the pairs $F^2$ with the given operations form a field, and there are $p$ choices for each coordinate, so its cardinality is $\boxed{p^2}$.

Suppose $x^2+y^2=0$ in the field $F$. If $y\ne 0$, division by $y^2$ gives $(x/y{)}^2=-1$, contrary to the hypothesis. Hence $y=0$ and then $x^2=0$; the absence of nonzero zero divisors in a field gives $x=0$. ThereforeFor the stated operations on $F^2$, addition is an abelian group operation: associativity and commutativity hold componentwise, its identity is $(0,0)$, and the negative of $(x,y)$ is $(-x,-y)$.

To check multiplication without omitting associativity, associate to each pair the matrixThis is an injection into the two-by-two matrices over $F$, and direct matrix multiplication givesIt also preserves addition. The associativity and distributivity of matrix multiplication therefore imply the same laws for the pair multiplication. Its formula is symmetric in the pairs, so it is commutative. Its identity is $(1,0)$, which is different from $(0,0)$.

For a nonzero pair, the first result shows that $x^2+y^2\ne 0$. Its multiplicative inverse issince multiplying $(x,y)$ by $(x,-y)$ gives $(x^2+y^2,0)$. This verifies every field axiom, so the given operations make $F^2$ a field. The embedded copy of $F$ is $\{(x,0):x\in F\}$; the element $(0,1)$ has square $(-1,0)$, describing a quadratic extension of $F$.

The greatest common divisor issince $19$ and $44$ are coprime integers. Part (b) therefore gives $\boxed{1012/23=44}$ distinct $437$th powers in the nonzero multiplicative group of a finite field ${\mathbb{F}}_{1013}^{\times }$. If “powers mod $1013$” includes all residue classes, the additional value $0=0^{437}$ gives $\boxed{45}$ values including zero. The distinction is relevant because part (b) explicitly works in the nonzero group.

Take a generator $u$ of the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$, which has order $p-1$. Write $x=u^a$. Since any possible nonzero base is $y=u^r$, the condition of being a $k$th power isBy part (a), this linear congruence is soluble exactly when $d=\gcd (k,p-1)$ divides $a$. MeanwhileThus the required test isThe exponent residue classes divisible by $d$ are $0,d,\dots ,p-1-d$, so there are exactly $(p-1)/d$ distinct nonzero $k$th powers. Equivalently, part (a) shows each value has exactly $d$ preimages under the power map.

Let $d=\gcd (a,m)$. If the linear congruence $ar\equiv b(\mathrm{mod}m)$ holds, then $b=ar-ms$ for some integer $s$, so the greatest common divisor $d$ divides $b$. Conversely, Bézout's identity gives integers $u,v$ with $au+mv=d$. If $b=dj$, multiply this identity by $j$ to obtain $a(uj)\equiv b(\mathrm{mod}m)$. HenceFor the number of solutions, write $a=d{a}^{\prime }$, $m=d{m}^{\prime }$ and $b=d{b}^{\prime }$. The linear congruence becomes $a^{\prime }r\equiv b^{\prime }(\mathrm{mod}{m}^{\prime })$, where $a^{\prime },m^{\prime }$ are coprime integers. Multiplication by $a^{\prime }$ is invertible modulo $m^{\prime }$ by Bézout's identity, so the solutions form one residue class $r\equiv r_0(\mathrm{mod}{m}^{\prime })$. Modulo $m=d{m}^{\prime }$, these are exactlyThey are distinct: if two have the same residue class modulo $m$, then $d{m}^{\prime }$ divides $(j-\ell )m^{\prime }$, forcing $d\mid j-\ell$; with $0\le j,\ell <d$, this forces $j=\ell$. Every other solution differs from $r_0$ by a multiple of $m^{\prime }$, so these exhaust the $d$ solutions.

Finally,Combining this with the existence criterion gives the requested equivalent modular congruence.

To obtain a countable closure under real polynomial roots, start with $B_0=\{1,\pi \}$ and explicitly retain previous elements at each stage:Part (b) shows that $\varphi$ takes countable sets to countable sets; consequently all $B_n$ and then $X$ are countable by the countable union of countable sets theorem. Also $1,\pi \in X$.

For a nonzero polynomial $P(t)=c_0+\cdots +c_d{t}^d$ with all $c_j\in X$, choose stages $n_j$ containing the finitely many coefficients and put $N={\max }_j{n}_j$. The inclusions $B_n\subseteq B_{n+1}$ ensure that every coefficient belongs to $B_N$. Every real root of a polynomial $P$ therefore belongs to $\varphi (B_N)\subseteq B_{N+1}\subseteq X$. This countable set $X$ has all the required properties. Retaining $B_n$ explicitly avoids assuming the generally false inclusion $A\subseteq \varphi (A)$: for example, $\varphi (\{1\})$ does not contain $1$, since a nonzero polynomial all of whose coefficients are one is positive at $1$.