For the regular Sturm-Liouville setting assume $p$ is positive and continuously differentiable, $q,w$ are continuous, and $w>0$. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation, divided by two, isEndpoint-vanishing variations and the allowed fundamental lemma giveIf an endpoint value is free, its variation is arbitrary, giving the natural condition $p{y}^{\prime }=0$ at that endpoint. At a prescribed endpoint value the variation vanishes there, so no additional natural condition is imposed.

For eigenfunctions $u,v$ with different eigenvalues, multiply their equations by $v,u$ and subtract after integration. The resulting boundary expression $[p(u{v}^{\prime }-u^{\prime }v){]}_a^b$ vanishes under common homogeneous Dirichlet or common natural boundary conditions. Hence $({\lambda }_u-{\lambda }_v){\int }_a^{b}wuv=0$, proving weighted orthogonality.

For the final Dirichlet problem the normalized stationary modes areSolving $y^{\prime \prime }+(\lambda +1)y=0$ with both endpoint values zero gives these and excludes nonpositive $\lambda +1$. Distinct sine modes have zero integral product on $[0,1]$. Their Dirichlet values also make the same boundary expression vanish, which explains orthogonality even though their endpoint derivatives do not obey the natural conditions.

For global minimality, the Fourier sine expansion of an endpoint-vanishing function and Parseval's identities give $\int y^{\prime 2}=\sum k^2{\pi }^2{a}_k^2\ge {\pi }^2\sum a_k^2={\pi }^2\int y^2$. Equality holds exactly in the first sine eigenspace. ThereforeDistinct modes are orthogonal; the two sign choices of the same minimum are not orthogonal to each other. This distinguishes the family of stationary modes from the two representatives of the minimizing eigenspace in the question's wording.

The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all solve $X^d=1$. By the allowed root bound they are all its solutions. A residue $b$ of order $d$ must therefore equal $a^j$; conversely $\mathrm{ord}(a^j)=d/\gcd (d,j)$, because $d\mid jt$ characterizes the exponents $t$ with $(a^j{)}^t=1$. This proves the equivalence with $\gcd (d,j)=1$ and gives exactly $\phi (d)$ elements of order $d$ whenever any such element exists.

To establish existence at the full order too, put $m=p-1$ and let $N_d$ count elements of order $d$. By Lagrange's theorem for finite groups, these orders divide $m$. The first result gives $N_d=0$ or $\phi (d)$. Thereforeusing the standard divisor-sum identity for Euler's function. Equality forces every deficit to vanish, so $N_m=\boxed{\phi (p-1)}$.

Finally let $h=g^{(p-1)/8}$, which has order eight. Its fourth power is the unique nonidentity square root of one, namely $-1$. Since $h^7=h^{-1}$,because $h^{-2}=h^6=-h^2$. This proves the stated residue squares to two.

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.