Set $A(y)=y^{-1/2}+\lambda$ and $F(y,y^{\prime })=A(y)\sqrt{1+y^{\prime 2}}$. Since there is no explicit $x$ dependence, the Beltrami identity says $F-y^{\prime }{F}_{y^{\prime }}$ is constant along a stationary curve. One can verify it directly:by the Euler-Lagrange equation. Here $F_{y^{\prime }}=A(y)y^{\prime }/\sqrt{1+y^{\prime 2}}$, soThis identity applies on the open arch where $y>0$. The constant is finite by evaluating at any interior point. Near an endpoint $A(y)>0$, hence $H>0$ everywhere on such an arch. Thus also $A(y)>0$ throughout it.

Let $g>0$ be the gravitational acceleration, and let $y$ denote positive downward depth. For a frictionless particle released from rest, conservation of energy gives $v^2=2gy$. The arc length element of $(x,-y(x))$ is $ds=\sqrt{1+y^{\prime 2}}dx$, so the travel time and length areFor a finite-time single arch, require $y>0$ in $0<x<1$, $y(0)=y(1)=0$, and interpret the endpoint integrals improperly. At a regular constrained minimizer with $S[y]=L$, the Lagrange multiplier rule for an isoperimetric constraint makes $\sqrt{2g}T+\lambda (S-L)$ stationary. Dropping the constant $-\lambda L$ givesFor $L>1$ the constraint is regular under interior variations: if its first variation vanished for every compactly supported variation, the Euler-Lagrange equation for $S$ would force $y^{\prime }$ constant, hence $y\equiv 0$ and $S=1$, a contradiction. Thus the multiplier is justified for a nondegenerate arch.

The printed case $L=1$ is degenerate. Since $\sqrt{1+y^{\prime 2}}\ge 1$, equality of the length with the horizontal separation forces $y^{\prime }=0$ and $y=0$. A particle initially at rest then has no tangential gravitational acceleration and cannot complete the trip. Its travel time is infinite. The finite-time calculation therefore concerns $L>1$, with endpoint regularity qualified below.

Fix $x\in U$. Since $U$ is open, choose an open ball $B(x,\epsilon )\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the chain rule givesThe ordinary mean value theorem therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a locally constant function.

For each $z\in {\mathbb{R}}^m$, its preimage is a union of such open balls, so $G^{-1}(\{z\})$ is open in ${\mathbb{R}}^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and connected, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. ThereforeThe empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.

Conjugate by a projective transformation sending $P$ to $[1:0]$ and $Q$ to $[0:1]$. A matrix representing the conjugated transformation must have zero diagonal, since its first column represents $[0:1]$ and its second represents $[1:0]$. Therefore it has the formIts square is $abI$, a nonzero scalar matrix, which induces the identity on the projective line. Conjugating back provesFor the requested explicit involution, takeMore generally $[x:y]\mapsto [ay:bx]$ works whenever $ab\ne 0$.

Take $p=5$. The unit $3$ modulo $121$ has order $5$, becauseLet a generator $b$ of $C_{25}$ act on $C_{121}=⟨a⟩$ by $a\mapsto a^3$. The action is well defined since its order $5$ divides $25$. The semidirect producthas order $121\cdot 25=11^2{5}^2$. Its construction as a semidirect product ensures that $a$ retains order $121$. Since $a^3\ne a$, the groups generated by $a$ and $b$ do not commute, so $G$ is nonabelian.

Apply the Abelian group criterion for a product of two prime squares with the smaller and larger primes in their correct order. For $p<11$, all such groups are Abelian exactly when $p\nmid 11^2-1=120$. Of the primes $2,3,5,7$, only $7$ qualifies. For $p>11$, the criterion is $11\nmid p^2-1$, equivalently $p\not\equiv \pm 1(\mathrm{mod}11)$. Therefore the exact answer isEquivalently, among the stipulated primes $p\ne 11$, exclude $2,3,5$ and the residue classes $\pm 1$ modulo $11$.

For completeness, the necessity in that criterion supplies actual counterexamples at every excluded prime. If $\ell <q$ and $\ell \mid q^2-1$, the multiplicative group of a finite field is cyclic, so choose $\alpha \in {\mathbb{F}}_{q^2}^{\times }$ of order $\ell$. Let a generator of $C_{{\ell }^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by multiplication by $\alpha$. This defines a nontrivial group action, hence a nonabelian semidirect product of order ${\ell }^2{q}^2$. Use $(\ell ,q)=(p,11)$ for $p=2,3,5$, and $(\ell ,q)=(11,p)$ for $p\equiv \pm 1(\mathrm{mod}11)$ with $p>11$. Thus no excluded prime can be restored to the answer.