For an odd prime $p$, Euler's criterion statesFor $p\nmid a$, the Legendre symbol is $1$ for a quadratic residue and $-1$ for a quadratic nonresidue; if $p\mid a$, both sides are zero.

The product is a single power because the arithmetic progression of exponents has sum $2017\cdot 2018/2$:Since $2017\equiv 1(\mathrm{mod}8)$, the second supplementary law for quadratic reciprocity gives $(2/2017)=1$. Hence Euler's criterion gives $2^{1008}\equiv 1(\mathrm{mod}2017)$. Reduce the exponent using $1009=1008+1$ and $2017=2\cdot 1008+1$:The least nonnegative remainder is therefore $2$.

For distinct odd primes $p,q$, quadratic reciprocity statesHere the Legendre symbol records whether a nonzero congruence class is a quadratic residue. The first supplementary law for quadratic reciprocity and second supplementary law for quadratic reciprocity giveThus minus one is a square exactly when $p\equiv 1(\mathrm{mod}4)$, and two is a square exactly when $p\equiv 1$ or $7(\mathrm{mod}8)$.

To test the specified congruence class, factor $111=3\cdot 37$. Since $2017\equiv 1(\mathrm{mod}4)$, multiplicativity of the Legendre symbol and quadratic reciprocity yieldThere is no sign change on swapping $19$ and $37$ because $37\equiv 1(\mathrm{mod}4)$. Finally $18=2\cdot 3^2$, so multiplicativity of the Legendre symbol gives $(18/19)=(2/19)=-1$, since $19\equiv 3(\mathrm{mod}8)$. Therefore

For $N\ge 2$, each adjacent pair of walkways has relative rapidity ${\varphi }_N=\mathrm{artanh}(1/N)$. There are $N$ such steps between $P_0$ and $P_N$, so the equal-rapidity chain of collinear Lorentz boosts has total rapidity $N{\varphi }_N$. Thus the actual velocity isIt is strictly below the speed of light for every finite $N\ge 2$. Adding the adjacent velocities as ordinary numbers ignores the relativistic velocity-addition formula.

For large $N$, the Taylor series of the inverse hyperbolic tangent givesConsequentlyMore precisely, $v_N/c=\tanh 1+{\mathrm{sech}}^2(1)/(3N^2)+O(N^{-4})$, so the limit is approached from above. The case $N=1$ would prescribe an adjacent relative velocity equal to the speed of light; it does not define an admissible walkway inertial frame.

Use coordinates $X=(ct,x,y,z{)}^T$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real $4\times 4$ matrix $L$ is a Lorentz transformation precisely when it preserves the Minkowski interval for every $X$, equivalentlyIt is proper when $\boxed{\det L=+1}$, and orthochronous when $\boxed{L^0{}_0>0}$, equivalently $L^0{}_0\ge 1$. The latter selects the orthochronous Lorentz group, which preserves future-directed timelike four-vectors. Indeed the time-time entry of the preservation identity gives $(L^0{}_0{)}^2-{\sum }_{i=1}^3(L^i{}_0{)}^2=1$, so $|L^0{}_0|\ge 1$. Imposing both conditions gives a proper orthochronous Lorentz transformation.

When an endpoint's horizontal coordinate is fixed but its height is free, the natural boundary conditions for a free endpoint require $F_{y^{\prime }}=0$ there. In this problem,Every admissible extremal is an upper circular arc with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the natural boundary conditions for a free endpoint at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, soIts free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole interval follows because the radicand is at least $4c^2$.

There can be no solution satisfying the natural boundary conditions for a free endpoint at both ends: an upper circular arc has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore

For the specified functional,Since $y>0$, the Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$ for some $R>0$. This suggests a circular arc, but it is worth deriving its equation without dividing by $y^{\prime }$: such division would lose the point where the arc is horizontal.

Substitute the displayed partial derivative and $F_y=-\sqrt{1+y^{\prime 2}}/y^2$ in the full Euler-Lagrange equation. Combining terms over the common denominator givesThus $(y{y}^{\prime }{)}^{\prime }=-1$, so $y{y}^{\prime }=h-x$ and a second integration givesThis establishes the semicircular extremals of the hyperbolic length functional and excludes constant $y$, which would satisfy the first-order Beltrami identity for a suitable $R$ but not the full Euler-Lagrange equation.

The two prescribed endpoints giveSubtracting yields $4h=4(d^2-c^2)$. ThereforeIn particular the centre is $\boxed{(d^2-c^2,0)}$. Admissibility requires $c,d>0$, as the prescribed heights must lie in $y>0$. Under that condition the radicand is positive throughout $[-1,1]$: it is a concave quadratic polynomial whose two endpoint values are $4c^2$ and $4d^2$. The positive root consequently gives one smooth admissible arc, uniquely determined by the endpoints.

Use the chain rule along a sufficiently smooth extremal of the functional. For $E=y^{\prime }{F}_{y^{\prime }}-F$,The bracket vanishes by the Euler-Lagrange equation. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the Beltrami identity:

Put $G(p)=\mathcal{L}[J_0](p)$, initially for real $p>0$. Besides the Laplace transform of a derivative, use multiplication by the original variable in the formTransforming the Bessel differential equation, with its given initial conditions, yieldsThe constant terms cancel, leaving $(1+p^2)G^{\prime }+pG=0$. This first-order ordinary differential equation integrates toThe initial value theorem for Laplace transforms gives ${\lim }_{p\to \infty }pG(p)=J_0(0)=1$, so $C=1$. Thus the Laplace transform of the Bessel function J0 isFor complex $p$ with $\mathrm{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.

For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2{)}^{-3/2}$. On the other hand, the same multiplication rule givesTaking an integral from zero divides the Laplace transform by $p$. By uniqueness of the Laplace transform,A direct check also explains the cancellation. This expression has $f^{\prime }=x{J}_0$ and $f^{\prime \prime }=J_0+x{J}_0^{\prime }$. The Bessel differential equation says $(x{J}_0^{\prime }{)}^{\prime }=-x{J}_0$, so integration from zero gives $x{J}_0^{\prime }=-{\int }_0^{x}t{J}_0(t)dt=-f$. Hence $f^{\prime \prime }+f=J_0$, and both initial conditions hold.