Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2

For odd positive coprime integers, quadratic reciprocity for the Jacobi symbol isIf the integers are not coprime, both symbols vanish; equivalently the usual signed reciprocity identity still holds. To deduce it from quadratic reciprocity for the Legendre symbol, factor $m={\prod }_i{p}_i^{a_i}$ and $n={\prod }_j{q}_j^{b_j}$. Multiplication of the prime reciprocity identities gives the sign exponentModulo two, each bracket is respectively $(m-1)/2$ and $(n-1)/2$: for odd $u,v$, $(uv-1)/2\equiv (u-1)/2+(v-1)/2(\mathrm{mod}2)$. This proves the composite-denominator law, including repeated prime factors.

Now $261=9\cdot 29$, and 9 contributes a square factor. Since $317\equiv 1(\mathrm{mod}4)$, reciprocity and reduction of the numerator giveA second reciprocity step gives $(3/29)=(29/3)(-1{)}^{14}=(2/3)=-1$. Thus the requested Jacobi symbol is $-1$.

The Chebyshev alternation theorem says that for $f\in C([a,b])$ a polynomial $q$ of degree at most $n-1$ is a best uniform approximation precisely when there are $n+1$ ordered points $a\le x_0<\cdots <x_n\le b$ such thatThe best polynomial is unique. These equally large alternating extrema are the equal-ripple criterion; the zero-error case is interpreted in the evident way.

Set $p_{\ast }(x)=2^{1-n}{T}_n(x)$, a monic Chebyshev polynomial. Its norm on $[-1,1]$ is $2^{1-n}$. At the ordered points $x_j=\cos ((n-j)\pi /n)$, its values alternate between the two extreme values $\pm 2^{1-n}$.

Approximate $f(x)=x^n$ by polynomials of degree at most $n-1$. The error corresponding to $q_{\ast }=x^n-p_{\ast }$ is exactly $p_{\ast }$. The Chebyshev alternation theorem therefore makes $q_{\ast }$ the unique best approximant. For any monic $p$, the polynomial $q=x^n-p$ is an admissible approximant, soEquality holds precisely for $p=p_{\ast }$. Alternatively, a smaller norm would force $p-p_{\ast }$ to change sign between each consecutive pair of alternating extrema, giving at least $n$ zeros to a polynomial of degree at most $n-1$.

In the upper-half-plane model, with metric $d{s}^2=(d{x}^2+d{y}^2)/y^2$, the hyperbolic geodesics are vertical lines and semicircles orthogonal to the real boundary. Two geodesics with disjoint endpoints and no intersection are ultraparallel and have a common perpendicular $h$. To construct it, send one geodesic to the imaginary axis by an isometry. The second then has endpoints $0<a<b$ after a possible reflection. The semicircle centered at zero with radius $\sqrt{ab}$ is perpendicular to the imaginary axis and to the second circle: its squared radius plus $((b-a)/2{)}^2$ equals $((a+b)/2{)}^2$, the orthogonality condition for the two circles.

Let the intersection points of $h$ with the given geodesics have signed arclength coordinates $s_1,s_2$ on $h$. Each hyperbolic reflection preserves $h$ and restricts there to $s\mapsto 2s_i-s$: the perpendicular intersection point is fixed and the tangent direction along $h$ is reversed. Thus their composition restricts toThe displacement is nonzero since the two perpendicular intersection points differ. Its $N$th iterate translates by $2N(s_1-s_2)$, which is never zero for $N>0$. The composite is a hyperbolic translation of infinite order. The assumption excluding a common endpoint ensures the common perpendicular used here exists.

The supplied congruence is a congruence of squares, since $25=5^2$. HenceA nontrivial square root modulo a product of two odd primes may agree with $+5$ modulo one prime and $-5$ modulo the other. Taking greatest common divisors separates these factors. The Euclidean algorithm givesFor example the successive remainders for the first computation are $1072,737,335,67,0$; for the second they are $1062,767,295,177,118,59,0$. ThereforeBoth numbers are prime by trial division by primes at most their square roots. More generally, given $a^2\equiv b^2(\mathrm{mod}N)$ with $a\not\equiv \pm b(\mathrm{mod}N)$, computing $\gcd (N,a-b)$ often produces a proper factor. The method requires no prior knowledge of which prime gives which sign, and is the basic factor-extraction step in congruence-of-squares algorithms.

An exponential dispersion family has density or mass functionwith support independent of the natural parameter $\theta$. Usually $a(\varphi )=\varphi /w$ for a known weight $w$. For the Poisson distribution, identifyThe dispersion is fixed at one. Differentiating the normalizing identity gives $\mathbb{E}Y=b^{\prime }(\theta )$; differentiating again gives $\mathrm{Var}Y=a(\varphi )b^{\prime \prime }(\theta )$. Here both derivatives equal $e^{\theta }=\lambda$, soThe variance function expresses $b^{\prime \prime }(\theta )$ in terms of the mean $\mu$. The canonical link function maps the mean to the natural parameter, so the Poisson canonical link is $g(\mu )=\log \mu$.

By the law of mass action, the net changes in the intermediate concentrations areThe autocatalytic reaction increases $X$ by one and decreases $Y$ by one; its reaction-rate convention is $k_3{X}^{2}Y$. With the specified common concentration scale and time scale, chooseSubstitution gives the nondimensional Brusselator system. Adding its two steady equations gives $u=1$, and then the second gives $v=b/a$. Its Jacobian matrix there isThe equilibrium is asymptotically stable for $b<1+a$ and unstable for $b>1+a$. At $b_c=1+a$ the eigenvalues are $\pm i\sqrt{a}$, crossing the imaginary axis with nonzero speed $\frac{d}{db}\mathrm{Re}\lambda =1/2$.

The nonlinear Hopf bifurcation is nondegenerate. At the threshold, put $x=u-1$ and $y=-\sqrt{a}[(u-1)+(v-b/a)]$. The translated equations areThe Hopf coefficient of the Brusselator in the planar radial normalization is $-(a+2)/8<0$: the cubic contribution is $-6a/16$ and the quadratic contribution is $[-2\sqrt{a}2(1-a)]/(16\sqrt{a})$. Consequently the bifurcation is supercritical, producing small stable oscillations on the unstable-equilibrium side. Their limiting period is

Write $F(x)={\mu }^2-(x^2-a{)}^2$. Its fixed points satisfy $x^2=a\pm |\mu |$, retaining only nonnegative right sides. At a simple fixed point,Thus on the outer branches $x^2=a+|\mu |$, the positive point is stable and the negative point unstable. On the inner branches $x^2=a-|\mu |$, the positive point is unstable and the negative point stable.

For $a>0$, there are four simple points for $0<|\mu |<a$, two for $|\mu |>a$, and a saddle-node at each $(\mu ,x)=(\pm a,0)$, where the two inner points coalesce. At $\mu =0$, there are also bifurcations at $x=\pm \sqrt{a}$. The smooth branches $x=\pm \sqrt{a+\mu }$ and $x=\pm \sqrt{a-\mu }$ cross and exchange stability. Locally $F={\mu }^2-4a{\xi }^2+\text{higher terms}$, where $\xi =x-x_0$. After a parameter-dependent shift this is a transcritical bifurcation. It is not an ordinary saddle-node: two nearby branches exist on both sides of zero. At zero itself each crossing point is semistable, attracting from the right and repelling from the left.

For $a<0$, the only possible branch is $x=\pm \sqrt{a+|\mu |}$, existing for $|\mu |\ge |a|$. Each point $(\pm |a|,0)$ is a saddle-node bifurcation. There are no fixed points for $|\mu |<|a|$; outside that interval the positive point is stable and the negative point unstable. At either threshold $F\sim 2a{x}^2<0$, so the zero point is semistable. The first row of the diagram displays these branches; solid curves are stable and dashed curves unstable.

Now $G(x)=x[{\mu }^2-(x^2-a{)}^2]$. The zero point exists for all parameters and has eigenvalue $G^{\prime }(0)={\mu }^2-a^2$, so it is stable for $|\mu |<|a|$ and unstable for $|\mu |>|a|$. At every nonzero fixed point,Both outer branches are stable and both inner branches unstable.

For $a>0$, the nonzero locations are as in part (i). At $(\mu ,x)=(\pm a,0)$ the local equation is $\dot{x}=({\mu }^2-a^2)x+2a{x}^3+O(x^5)$: these are subcritical pitchfork bifurcations, with the two unstable inner branches present on the side where zero is stable. At the bifurcation itself $\dot{x}\sim 2a{x}^3$, so zero is unstable. At $(0,\pm \sqrt{a})$ the inner and outer branches cross and exchange stability in transcritical bifurcations. The positive crossing is semistable, attracting from the right; the negative crossing is semistable, attracting from the left, because the additional factor $x$ reverses the local flow there.

For $a<0$, the outer stable pair exists only for $|\mu |>|a|$. The points $(\pm |a|,0)$ are supercritical pitchfork bifurcations, since the cubic coefficient $2a$ is negative. Zero loses stability as $|\mu |$ increases across $|a|$, and a stable symmetric pair appears. At the threshold zero remains nonlinearly stable, with $\dot{x}\sim 2a{x}^3$. There are no further bifurcations. The second row of the preceding figure gives the diagrams. The plots use $a=\pm 1$; the general locations are obtained from the formulas above.

Use the sign convention in the question: the Hilbert transform kernel is $(y-x{)}^{-1}$, the negative of another common convention. A contour indentation at the origin givesFor $s>0$ close the contour in the upper half-plane; for $s<0$ close it below. Thus the transform of $\sin (sx)$ is $\cos (sx)$ for $s>0$. Since $(1-\cos x)/x={\int }_0^1\sin (sx)ds$, integration of this multiplier identity, justified first with an Abel regularization, yieldsThe value at zero also follows directly from ${\pi }^{-1}\int (1-\cos y)y^{-2}dy=1$.

Splitting the exponential into the half-plane where it decays givesTheir singularities at zero are removable. The first function is analytic above the real axis and the second below it; their exponential terms are bounded in those respective half-planes, giving the required $O(1/z)$ decay. Their difference on the real axis is the prescribed jump. Moreover their sum there is $-i\sin x/x$, agreeing with the Sokhotski–Plemelj formula and the transform computed above.

For uniqueness in the class with continuous boundary values, subtract two solutions. The zero jump glues their differences into an entire function, tending to zero at infinity. Liouville's theorem makes it zero. Thus these are the unique solutions of the Riemann-Hilbert problem with the stated decay.

For $q=(x_1,{\eta }_1,{\eta }_2{)}^T$, the Euler-Lagrange equations giveIn components these are $m{\ddot{x}}_1=-\mu (19x_1/3-{\eta }_1)$, $m{\ddot{\eta }}_1=-\mu (2{\eta }_1-x_1-{\eta }_2)$ and $m{\ddot{\eta }}_2=-\mu ({\eta }_2-{\eta }_1)$. A normal mode has $q=v\cos (\omega t+\delta )$ and satisfies $Kv=(m{\omega }^2/\mu )v$. For eigenvalue $1/3$, the first and third rows give ${\eta }_1=6x_1$ and ${\eta }_2=3{\eta }_1/2$, and the second row then agrees. HenceThe particle positions are recovered by adding the equilibrium offsets to the last two entries. Their displacements are in phase with amplitude ratio $1:6:9$.

Substitute $x=pc/(k_{B}T)$ in the Fermi-Dirac distribution energy integral. ThenThe integral converges, is positive, and is independent of temperature. Thus $ϵ=\alpha T^4$, with $\alpha$ equal to the displayed constant integral. The given normalization is for the specified degeneracy; further internal degrees of freedom multiply it.

For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.

The planar Brouwer fixed-point theorem states that every continuous map from a nonempty compact convex subset of ${\mathbb{R}}^2$ to itself has a fixed point. In particular it applies to a closed disk or a closed triangle. Convexity and compactness are essential hypotheses in this formulation.

Let $v_I,v_J,v_K$ be the vertices opposite the corresponding sides, and let ${\lambda }_I,{\lambda }_J,{\lambda }_K$ be their barycentric coordinates. We prove the equivalence by negating each assertion.

Suppose the closed covering sets have empty triple intersection. For $x\in T$ put $d_I(x)=\mathrm{dist}(x,A)$, and define $d_J,d_K$ similarly. At least one distance is positive, since $x$ cannot belong to all three sets, and at least one is zero, since they cover $T$. Consequentlyis continuous and lies in the boundary: one barycentric coordinate vanishes. On $I$ its first coefficient is zero because $I\subseteq A$, so $f(I)\subseteq I$, and likewise for the other sides. This contradicts assertion (1), proving (1) implies (2).

Conversely, suppose such a boundary-valued map exists. Define closed subsets of ${\mathbb{R}}^2$ byThey are closed because $T$ is compact and the coordinate functions are continuous. They cover $T$ because $f$ is boundary-valued; the side-preserving condition gives $I\subseteq A$, $J\subseteq B$, $K\subseteq C$. Their triple intersection is empty because barycentric coordinates sum to one. Thus failure of (1) implies failure of (2), completing the equivalence.

In fact both assertions hold. If the map in (1) existed, let $R$ rotate the equilateral triangle through $120^{\circ }$. Brouwer fixed-point theorem applied to $R\circ f:T\to T$ gives a fixed point $x$ on the boundary. On any side $I$ containing $x$, side preservation puts $x$ also on $R(I)$, so it is their common vertex. A side-preserving map fixes every vertex because that vertex lies on two sides. Therefore $x=Rf(x)=Rx$, impossible for a noncentral vertex. This also proves the covering-intersection assertion through the established equivalence.

On the unit circle the two equations reduce to the same conditionsince $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y{)}^2+g(x,y{)}^2)-f(x,y{)}^2$. At either point $(\pm 1,0)$ it is strictly positive, and at either point $(0,\pm 1)$ it is strictly negative. On each of the four closed quarter-circle arcs, continuity and the intermediate value theorem give a zero in its interior. These four interiors are disjoint, so there are at least four distinct solutions, one in each open quadrant. No solution here lies on an axis because $f,g$ are strictly positive.

The standard name is the Reed-Muller code. For $0\le d\le m$, evaluate all multilinear polynomials over ${\mathbb{F}}_2$ of total degree at most $d$ at every point of ${\mathbb{F}}_2^m$. These evaluation vectors form $RM(m,d)$, of length $2^m$. Distinct multilinear polynomials have distinct evaluations: induct on the variables using $f=g+x_{m}h$ and the two restrictions $g,g+h$. Thus the monomials of degree at most $d$ are independent andFor minimum weight, use the same splitting. A nonzero $h$ has weight at least $2^{m-d}$ by induction, and $\mathrm{wt}(g)+\mathrm{wt}(g+h)\ge \mathrm{wt}(h)$. If $h=0$, the two identical halves give the same lower bound by induction on $g$. The endpoint $d=m$ has minimum weight one, and $d=0$ is the repetition code. A product of $d$ distinct variables has exactly $2^{m-d}$ nonzero evaluations, soHere distance equals minimum nonzero weight because the code is linear.

Every monomial of degree at most $d-1$ in $d$ variables has an even number of ones, and parity is additive over ${\mathbb{F}}_2$. Thus every word of $RM(d,d-1)$ has even weight. If $x\in RM(m,r)$ and $y\in RM(m,m-r-1)$, their coordinatewise product is an evaluation of degree at most $m-1$, even after reducing $X_i^2=X_i$. Its weight is therefore even, so $⟨x,y⟩=0$. This proves the asserted inclusion in the dual code. Finallyby binomial symmetry. The dimensions agree with that of the orthogonal complement, proving $RM(m,r{)}^{\perp }=RM(m,m-r-1)$ for $0\le r<m$.

For a positive-speed travelling wave, the nutrient equation gives $a^{\prime }=kb/c$. Integrating the bacterial equation once, with zero flux in the uncolonized limit, givesHenceIn the region where $b>0$, divide the two equations. With $\gamma =\chi /D$,The condition $b(1)=0$ fixes the integration constant, givingAt $\chi =D$ the limiting expression is $b=-(c^2/kD)a\log a$. These formulas are positive between zero and one. For $\chi <D$ the nutrient reaches zero at a finite rear coordinate, so the strictly positive smooth whole-line ansatz with logarithmic chemotactic sensitivity needs a separate zero-nutrient interpretation there; for $\chi \ge D$ it reaches zero only as $z\to -\infty$.

For $\chi =2D$, the nutrient equation reduces to $a^{\prime }=(c/D)a(1-a)$ and integration givesThe factor $K$ in the bacterial numerator is required by $a^{\prime }=kb/c$; it is absent in the printed expression unless $K=1$. A translation of $z$ sets $K=1$, which is the version plotted below. The nutrient rises monotonically from zero behind the band to one ahead, while bacteria form a localized pulse with maximum $c^2/(4kD)$ at the nutrient half-height. Its total population is $\int bdz=c/k$. The band advances into fresh nutrient, consumes it and leaves a depleted region behind; chemotaxis draws bacteria up the nutrient gradient while diffusion spreads them.

Use the half-line transform $\hat{u}(k,t)={\int }_0^{\infty }{e}^{-ikx}u(x,t)dx$ and let $g_j(t)={\partial }_x^{j}u(0,t)$. Integrating the heat equation twice by parts givesHere ${\hat{u}}_0(k)=(1+ik{)}^{-2}$ andwith removable singularities at the apparent denominator zeros. Let $D^{+}=\{\mathrm{Im}k>0,\mathrm{Re}{k}^2<0\}$, and orient its boundary from infinity on the $3\pi /4$ ray to zero and then out on the $\pi /4$ ray. Transform inversion and contour deformation giveTo eliminate the unknown boundary derivative, evaluate the transform identity at $-k$: ${\tilde{g}}_1={\hat{u}}_0(-k)+ik{\tilde{g}}_0-e^{k^{2}t}\hat{u}(-k,t)$. The resulting extra contour integral of $e^{ikx}\hat{u}(-k,t)$ vanishes by analyticity and decay in the upper half-plane. This proves that the boxed expression involves only the prescribed data. It is intended for $x,t>0$, with boundary and initial values obtained as limits.

Split the real contour at zero and rotate its positive part to angle $\theta \in (0,\pi /4)$ and its negative part to angle $\pi -\theta$. Rotate the two rays of $\partial D^{+}$ to these same angles, preserving orientation. The initial-transform pole at $k=i$ is not crossed by either real-contour rotation, while ${\hat{u}}_0(-k)$ has its pole at $-i$, outside the upper half-plane. The apparent poles in ${\tilde{g}}_0$ are removable.

On the rotated rays, both $\mathrm{Im}k>0$ and $\mathrm{Re}{k}^2>0$. The initial-data factor has Gaussian decay $|e^{ikx-k^{2}t}|=e^{-x\mathrm{Im}k-t\mathrm{Re}{k}^2}$. For the boundary term, retain the integral formwhich is $O(|k{|}^{-2})$ there. Multiplication by $e^{ikx}$ gives exponential decay as $|k|\to \infty$ for every fixed $x>0$. Thus all integrands are exponentially damped on the deformed contours; the endpoint boundary value at $x=0$ is taken afterwards, rather than claiming uniform exponential decay there.

The angles $\psi$ and $\varphi$ are cyclic. Their conjugate momentum components are $p_{\psi }=I_3(\dot{\psi }+\dot{\varphi }\cos \theta )$ and $p_{\varphi }=I_1\dot{\varphi }{\sin }^2\theta +p_{\psi }\cos \theta$, so both are conserved; consequently ${\omega }_3=p_{\psi }/I_3$ is constant. With $q=I_3{\omega }_3=p_{\varphi }>0$,Eliminating the cyclic velocities gives the nutation effective potentialapart from the constant axial spin energy. ThusThe endpoint $\theta =\pi$ is excluded for these nonzero conserved momenta: the effective potential diverges there. The apparent Euler-angle singularity at $\theta =0$ is interpreted by its smooth small-tilt limit.

Put $\beta =q^2/(4I_{1}gl)>1$. There is no nonzero equilibrium angle because ${\cos }^4(\theta /2)\le 1<\beta$. The upright angle $\theta =0$ is stable: its linearized equation is $\ddot{\theta }=-(gl/I_1)(\beta -1)\theta$. HenceThese are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.

For $0<\beta <1$, the upright angle has $\ddot{\theta }=(gl/I_1)(1-\beta )\theta$ and is unstable. There is one nonzero equilibrium in $0<\theta <\pi$,Differentiate the right side of the nutation equation at this angle. Since the bracket vanishes there, the derivative is $-2(gl/I_1)\sin {\theta }_{\ast }\tan ({\theta }_{\ast }/2)=-4(gl/I_1){\sin }^2({\theta }_{\ast }/2)$. Therefore it is stable, withThis equilibrium represents steady precession, not a top with all Euler angles stationary. The equality case $\beta =1$ is excluded by the two cases in the question and requires a nonlinear marginal analysis.

One Hilbert axiomatization of classical propositional logic uses the schemas $A\to (B\to A)$, $[A\to (B\to C)]\to [(A\to B)\to (A\to C)]$ and $(\neg B\to \neg A)\to (A\to B)$, with modus ponens as its inference rule. Other connectives can be defined from implication and negation. Syntactic entailment $\Gamma ⊢A$ means that a finite formal proof derives $A$ from the schemas and members of $\Gamma$. Consistency means that no formula and its negation are both derivable, equivalently no contradiction is derivable.

Use the standard deduction theorem, closure of derivability under modus ponens, and the maximal-consistency properties: a maximal consistent set contains exactly one of $A,\neg A$, is closed under derivability, and satisfies $A\to B\in \Phi$ exactly when $A\notin \Phi$ or $B\in \Phi$. These follow from the deduction theorem and the classical schemas; in particular adjoining a missing formula to a maximal set must yield a contradiction. Define $v(p)=T$ iff $p\in \Phi$. Induction on formula construction now provesFor negation use the exactly-one property, and for implication use its displayed membership rule, which is precisely the Boolean truth table. Thus every member of $\Phi$ is true under $v$, providing the required valuation.

Add to the group axioms the sentences $\forall x(x^n=e\Rightarrow x=e)$ for every $n\ge 2$. Their models are precisely the torsion-free groups. This theory is not finitely axiomatizable. If a finite sentence or finite theory were equivalent to it, the compactness theorem would imply that finitely many of these power axioms already entailed the proposed axiomatization. A cyclic group of prime order greater than all the finitely listed exponents satisfies those power axioms but is not torsion-free, a contradiction.

The class of all torsion groups is not first-order axiomatizable at all. If a theory $T$ axiomatized it, add a new constant $c$ and the sentences $c^n\ne e$ for all $n\ge 1$. Every finite subset has a model: use a cyclic group of sufficiently large prime order, with $c$ a generator. Compactness gives a model of all of $T$ in which $c$ has infinite order, contradicting the claimed torsion property. These conclusions concern the ordinary first-order language of groups.

The Turán graph $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one. Turan theorem states that a graph on $n$ vertices containing no $K_r$ has at most $e(T_{r-1}(n))$ edges, with equality precisely for that balanced complete multipartite graph.

Here is a symmetrization proof, including equality. Choose a $K_r$-free graph with the maximum number of edges. Replacing a vertex $u$ by a nonadjacent twin of a nonadjacent vertex $v$ preserves $K_r$-freeness: a clique using the replacement would give the same clique using $v$. Thus nonadjacent vertices must have equal degrees, or cloning the higher-degree vertex increases the edge count. In fact their neighborhoods are identical. Otherwise choose $w$ adjacent to $u$ but not $v$. Cloning $u$ to $v$ preserves the total number of edges but lowers $w$'s degree by one, leaving $v$'s degree unchanged. The still nonadjacent pair $w,v$ now has unequal degrees, allowing an edge-increasing clone and contradicting maximality.