The integral Gauss's law isThe divergence theorem makes the left side ${\int }_V\nabla \cdot \mathbf{E}dV$. If the differential relation $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ holds, this proves the integral law. Conversely, if the integral identity holds for every suitable volume and the fields are sufficiently regular, a nonzero value of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ would persist in sign on a small neighborhood and contradict its zero integral. The differential law follows.

For a sphere of radius $r$, spherical symmetry makes the flux $4\pi r^{2}E(r)$ and the enclosed charge $4\pi {\int }_0^r{s}^2\rho (s)ds$. HenceThere is no extra point charge at the origin in the stipulated density distribution. Outside the charge support,

If $L$ is independent of $q_1$, the Euler-Lagrange equation makes the generalized momentum $p_1=\partial L/\partial {\dot{q}}_1$ constant. If $L$ is independent of time, the energy function $E={\sum }_a{\dot{q}}_a(\partial L/\partial {\dot{q}}_a)-L$ is constant; differentiating it gives $\dot{E}=-L_t$ along solutions.

Set $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$. Their derivatives are $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The radial and angular equations areEquivalently,Two conserved quantities areThe second equals $L$ here because the Lagrangian is homogeneous quadratic in velocities. For a constant-radius orbit with nonzero angular motion, the radial equation requires $r-3m=0$. ThusIt lies outside $2m$ and is the circular photon orbit. If both velocities were zero, the differential equations would also admit static constant configurations at any radius; these are degenerate zero-energy paths rather than the intended moving photon trajectory.

With fixed endpoint values, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=\text{constant}$.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so $y^{\prime \prime }=y$. The extremals arewhere the two fixed endpoint values determine $A,B$. The Beltrami constant is $y^2-y^{\prime 2}$, consistent with this solution.

For a nonzero smooth perturbation $\eta$ vanishing at the endpoints, expand the quadratic functional. The first variation vanishes at an extremal, whileThe linear term $\int y^{\prime }$ depends only on the fixed endpoints. Consequently the extremal is a strict minimum and cannot be a maximum.

The quadratic reciprocity states that for distinct odd primes $a,b$,For fixed odd prime $r$, first suppose that $p,q\ne r$. The congruence $p\equiv q(\mathrm{mod}4r)$ gives the same residues modulo $r$ and modulo four. Reciprocity yieldsIf $p=r$, the congruence forces $r\mid q$, so $q=r$ and both symbols are zero; the same handles $q=r$.

For $n={\prod }_j{r}_j^{a_j}$, where all $r_j$ are odd primes, $4r_j$ divides $4n$. Thus $p\equiv q(\mathrm{mod}4n)$ implies equality of each prime-numerator Legendre symbol. Multiplicativity givesThis also covers symbols that vanish; for $n=1$ both sides are one.

Eisenstein's criterion says that an integer polynomial with leading coefficient not divisible by a prime $\ell$, all other coefficients divisible by $\ell$, and constant coefficient not divisible by ${\ell }^2$ is irreducible over the rationals, provided it is primitive; the usual Gauss's lemma supplies the passage from integer to rational factors.

Apply it with $\ell =2$ toAll nonleading coefficients are even and the constant is not divisible by four. A translation of the variable preserves reducibility, so $x^8+1$ is irreducible over $\mathbb{Q}$.

Let $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. The minimal polynomial of $\zeta$ is $x^8+1$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Since $2c=\zeta +{\zeta }^{-1}$, the number $\zeta$ satisfies $t^2-2ct+1=0$ over $\mathbb{Q}(c)$. This real field cannot contain the nonreal $\zeta$, so that extension has degree exactly two. The tower law for field extensions gives

The hypothesis is a two-transitive group action. Given $\alpha ,\beta \ne {\omega }_1$, apply it to $({\omega }_1,\alpha )$ and $({\omega }_1,\beta )$. The resulting element fixes ${\omega }_1$ and sends $\alpha$ to $\beta$, so the stabilizer subgroup acts transitively on the remaining $n-1$ points.

The action on all of $\Omega$ is also transitive. By the orbit-stabilizer theorem,Thus $\boxed{n(n-1)\mid |G|}$. The diagonal pairs form one orbit, while two-transitivity makes all ordered unequal pairs a second orbit. Equality is preserved by a group action, so these cannot mix. There are exactly two orbits on $\Omega \times \Omega$.