In vacuum the source-free equations areAdding the divergence equations with weights $1,ic$ gives the complex divergence equation. Adding the curl equations in the same way gives $\nabla \times \mathbf{W}=(i/c){\mathbf{W}}_t$. Conversely real and imaginary parts of these two complex equations recover all four real equations.

For the cosine ansatz, differentiation yields divergence $(\mathbf{\Omega }\cdot \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$ and curl $-(\mathbf{\Omega }\times \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. The time derivative is $-\omega \mathbf{\Omega }\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. Since that sine is not identically zero, the required conditions are preciselyDotting the second equation with $\mathbf{\Omega }$ gives $\mathbf{\Omega }\cdot \mathbf{\Omega }=0$ because $\omega \ne 0$. This dot product is complex bilinear, not Hermitian. Taking the cross product again with $\mathbf{K}$ and using transversality gives $-|\mathbf{K}{|}^2\mathbf{\Omega }=-{\omega }^2\mathbf{\Omega }/c^2$. Nonzero amplitude therefore requires $\boxed{|\mathbf{K}{|}^2={\omega }^2/c^2}$.

Taking real and imaginary parts yields the full conditionsThey imply $\mathbf{e}\cdot \mathbf{b}=0$ and $|\mathbf{e}|=c|\mathbf{b}|$, together with the dispersion relation. Conversely a nonzero transverse $\mathbf{e}$, this dispersion relation and $\mathbf{b}=\mathbf{K}\times \mathbf{e}/\omega$ satisfy all the conditions.

The physical fields are $\mathbf{E}=\mathbf{e}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$ and $\mathbf{B}=\mathbf{b}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$. Each oscillates along one fixed real direction, so both fields are linearly polarized. Complex encoding of the two fields here does not mean either real field is circularly polarized.

Assume the regular Sturm–Liouville setting: real coefficients with $p>0$, $w>0$ and sufficient continuity and differentiability for the integrations below. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation in the direction $\eta$ isVariations vanishing at both endpoints and the supplied fundamental lemma giveAt a free endpoint the value of $\eta$ is arbitrary, so the natural condition is $p{y}^{\prime }=0$ there. At a fixed endpoint $\eta =0$, so no such derivative condition is imposed; the prescribed endpoint value is imposed instead. This distinction also allows one endpoint of each type.

For eigenfunctions $u,v$ of distinct eigenvalues $\lambda ,\mu$ under the same homogeneous self-adjoint endpoint conditions, multiply their equations by $v,u$ and subtract. Integration givesThe boundary expression vanishes for both natural conditions and homogeneous fixed-value conditions. Thus $\boxed{{\int }_a^{b}wuv=0}$ for different eigenvalues.

In the specified fixed-endpoint case the eigenproblem is $-y^{\prime \prime }-y=\lambda y$, $y(0)=y(1)=0$. There are no nonzero solutions when $\lambda +1\le 0$; the elementary exponential or affine solutions and the two endpoint conditions prove this. For $\lambda +1>0$, the first condition leaves a sine and the second requires frequency $n\pi$. The normalized stationary functions and eigenvalues areDistinct sine modes are orthogonal by the product-to-sum formula, or by the preceding Green second identity. Although they need not satisfy the natural derivative conditions, their fixed zero values also annihilate the boundary expression, so orthogonality is expected.

The Wirtinger inequality ${\int }_0^1{y}^{\prime 2}\ge {\pi }^2{\int }_0^1{y}^2$ gives $J[y]\ge {\pi }^2-1$. For completeness it follows by writing $y=v\sin \pi x$ for a smooth function supported away from the endpoints and integrating the cross term: $\int (y^{\prime 2}-{\pi }^2{y}^2)=\int {\sin }^2(\pi x)v^{\prime 2}\ge 0$. Approximation extends it to the Dirichlet Sobolev class. Equality is attained by the first sine mode; the eigenproblem or equality condition gives only its scalar multiples. Hence the two normalized minimizers are $\boxed{y=\pm \sqrt{2}\sin \pi x}$ and the minimum is $\boxed{{\pi }^2-1={\lambda }_1}$.

There is a wording distinction in the source: the two sign choices of the minimum are not orthogonal to each other, since their inner product is $-1$. Orthogonality applies to the stationary modes with different eigenvalues, not to both signs of the same minimizer.

The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all satisfy $X^d=1$. The supplied root bound says they exhaust those roots. Thus every element of order $d$ has the form $a^n$. If $g=\gcd (n,d)$, the least positive $k$ with $a^{nk}=1$ is $d/g$, soThis proves both directions of the first assertion.

For existence of a primitive root one may apply the finite-field cyclicity criterion proved in the preceding section. There is also a direct counting proof: for each divisor $d$ of $p-1$, either no elements have order $d$, or the preceding argument shows there are exactly $\varphi (d)$. Lagrange's theorem partitions all $p-1$ nonzero residue classes by such orders, while ${\sum }_{d\mid p-1}\varphi (d)=p-1$. Every count is at most its corresponding $\varphi (d)$, so equality of the totals forces equality at every divisor. In particular there are $\boxed{\varphi (p-1)}$ primitive roots.

Put $t=g^{(p-1)/8}$. It has order eight, so $t^4=-1$ and $t^8=1$. The specified element is $k=t+t^7=t+t^{-1}$. Since $t^{-2}=-t^2$, squaring gives $\boxed{k^2=t^2+2+t^{-2}=2(\mathrm{mod}p)}$.