Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4

Expand the finite Euler product into convergent geometric series. Every positive integer $n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, hence $P(x)\to 0$. If the sum of reciprocal primes were finite, then the inequality $-\log (1-t)\le 2t$ for $0\le t\le 1/2$ would giveuniformly in $x$, contradicting $P(x)\to 0$. Thus the sum of prime reciprocals diverges. Unique factorization justifies the expansion; no estimate for the distribution of primes is required. This is the Euler proof that the sum of reciprocals of primes diverges.

The closed-set form of the Baire category theorem says: if a nonempty complete metric space is a countable union of closed sets, at least one of those sets has nonempty interior. Equivalently, a complete metric space cannot be a countable union of closed sets with empty interiors. In the application the space is $\mathbb{R}$ with its usual complete metric.

Suppose the continuous functions converged pointwise to the rationality indicator. Each $F_N$ is closed, sincePointwise convergence implies that every point belongs to some $F_N$: its eventual values lie near either zero or one. Thus $\mathbb{R}={\bigcup }_N{F}_N$. By the Baire category theorem, some $F_N$ contains a nonempty open interval $I$. Choose a rational $r\in I$ and an irrational $s\in I$. For sufficiently large $n\ge N$, convergence gives $f_n(r)>2/3$ and $f_n(s)<1/3$. The intermediate value theorem then forces $f_n$ to take the value $1/2$ between them, at a point in $I\subseteq F_N$. This contradicts the defining condition of $F_N$. Therefore the rationality indicator is not a pointwise limit of continuous functions, the rationality indicator is not Baire class one result.

A plane crystallographic group is a discrete subgroup of the Euclidean plane's isometry group with compact quotient; equivalently a compact set has translates covering the plane. We prove the crystallographic restriction theorem without assuming its conclusion. The orientation-preserving subgroup $G^{+}$ has index at most two and still has a cocompact group action. Its subgroup $T$ of translations is discrete as an additive subgroup of ${\mathbb{R}}^2$.

Suppose $T$ were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in $G^{+}$ is a translation. Thus $G^{+}$ would be an abelian group. Every member would then commute with the given nontrivial rotation $g$ and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then $G$ would be finite, contradicting cocompactness on the unbounded plane. Hence $T$ contains a nonzero vector $v$.

If $g$ is a half-turn its order is already two. Otherwise its linear part $R$ rotates $v$ to a linearly independent vector $Rv$, and conjugating translations shows that $Rv\in T$. Thus $T$ spans the plane and is a Euclidean lattice. A discrete additive subgroup meets every bounded closed set in finitely many points: an accumulating sequence would give nonzero differences tending to zero, contradicting isolation of the identity. To see the Euclidean lattice assertion directly, choose a shortest nonzero translation $v$. Translations on its line are integer multiples of $v$. Reducing all parallel components modulo $v$ shows that nonzero perpendicular components are bounded away from zero, since otherwise infinitely many translation vectors would accumulate in a compact rectangle. Choose a vector $w$ with the smallest positive perpendicular component; reducing by $w$ and then $v$ proves $T=\mathbb{Z}v+\mathbb{Z}w$.

Conjugation by $g$ preserves this Euclidean lattice. In its Euclidean lattice basis, $R$ is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also $2\cos \theta$, where $\theta$ is the rotation angle. Since $g$ is nontrivial,These values give angles of order $2,3,4,6$ respectively. Hence

Shannon second coding theorem identifies the channel capacity of a discrete memoryless channel as $C={\max }_{P_X}I(X;Y)$: rates below $C$ admit block codes with error probability tending to zero, whereas a rate above $C$ cannot achieve vanishing error. Entropies here are measured in bits. For a binary erasure channel, a non-erased output determines its input exactly, while an erasure, independent of the input, leaves the original uncertainty unchanged. If $\mathbb{P}(X=1)=q$, thenThe binary entropy function has maximum one at $q=1/2$. ThereforeThis includes the noiseless case $p=0$ and the wholly erased case $p=1$.

The R command fits a Poisson regression with the default log link:with independent annual counts conditional on the supplied temperature index. The fitted mean is $\hat{\lambda }(T)=\exp (2.26061+0.48870T)$. The intercept gives about $9.59$ expected named storms at index zero. A one-unit increase in the index multiplies the expected count by $e^{0.48870}\simeq 1.63$; a $0.1$ increase multiplies it by about $1.050$.

The reported standard errors estimate the sampling uncertainty of the coefficients. Each displayed $z$ is the estimate divided by its standard error, compared with a standard normal distribution under a zero-coefficient null hypothesis. The slope has $z=2.879$ and two-sided $p=0.00399$, evidence of positive association within this model. The intercept test is against mean one at index zero and is not itself a test of a climatic mechanism. Association does not establish causation.

There are $61$ annual observations and two fitted coefficients, giving $59$ residual degrees of freedom. The residual deviance iswith $0\log 0=0$. It is not large relative to $59$ and gives no obvious overdispersion signal, although temporal dependence and omitted covariates still require substantive assessment.

For 2005 insert the index into the fitted mean:This is an expected count, approximately fourteen storms, rather than a guarantee of an integer outcome. In R one would use predict(Mod, newdata=data.frame(Temp=0.743), type="response"). A predictive distribution must include Poisson variation and, if desired, coefficient-estimation uncertainty; the latter needs the coefficient covariance matrix, not just the two printed standard errors.

Integrating the population equation over the line and using vanishing boundary flux gives $\dot{M}=-\mu M$, where $M(t)=\int n(x,t)dx$. Since $M(0)=Q$, one has $M(t)=Q{e}^{-\mu t}$. On the other hand the similarity substitution $x=\lambda (t)\eta$ givesConsequentlyThe decay condition is on the nonlinear diffusive flux $n{n}_x$; compact support also satisfies it.

For the similarity solution, differentiation givesChoose ${\lambda }^2\dot{\lambda }=e^{-\mu t}$. The remaining equation is $-(\eta f{)}^{\prime }=(f{f}^{\prime }{)}^{\prime }$. Its integration constant is zero for the required decaying or compactly supported profile, so $f(f^{\prime }+\eta )=0$. Where $f>0$, this gives $f^{\prime }=-\eta$. For a nonnegative population the complete profile is thereforeNormalization gives $Q=2b^3/3$, hence $b=(3Q/2{)}^{1/3}$. Integrating the scale equation with zero initial scale yieldsThusIts support satisfiesThe profile is a weak solution at its moving edges, where $n{n}_x$ vanishes. As $t\downarrow 0$ its support shrinks to the origin and its mass tends to $Q$, so its initial distribution is $Q{\delta }_0$. The bounded limiting range assumes $\mu >0$; at zero death rate the scale instead grows as $(3t{)}^{1/3}$.

The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta-gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.

For a uniform circular cross-section, averaging gives $⟨x^2⟩=⟨y^2⟩=R^2/4$; for a uniform axial coordinate in $[-h,h]$, $⟨z^2⟩=h^2/3$. Symmetry makes the coordinate axes principal axes, since all products of distinct coordinates integrate to zero. Thus the principal moments of inertia about the centre areThe inertia tensor is a scalar multiple of identity exactly when $h^2/3=R^2/4$. Thereforeis the height-to-radius ratio for sphere-like rotational inertia; the height-to-diameter ratio is $\sqrt{3}/2$. The geometric shape remains cylindrical, but its angular momentum is parallel to angular velocity for every axis, just as for a sphere.

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. HenceDifferentiating the Gaussian integral with respect to its coefficient gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. ThereforeThe degeneracy factor here is the one in the supplied integral, namely one. Substituting $kT=m{c}^2/\alpha$, $\mu =0$ and dividing by the supplied photon density givesAt $\alpha =20$ this is about $4.8\times 10^{-8}$. The relic is already nonrelativistic at decoupling and becomes still colder as its momenta redshift, making it a possible cold dark matter component if it is stable and interacts sufficiently weakly with light.

For orientation, write ${\eta }_b=n_b/n_{\gamma }$. Ignoring subsequent entropy transfer, its density relative to baryons would be ${\rho }_X/{\rho }_b=(1/20)(n/n_{\gamma })/{\eta }_b$. With the cosmological baryon-to-photon ratio of order $6\times 10^{-10}$, this is of order four, demonstrating the right cosmological scale. The measured dark-to-baryonic density ratio is likewise of order five; an observational reference is arxiv.org/abs/1807.06209 . This is a candidate abundance argument, not an exact relic-density calculation: photon heating after decoupling changes $n/n_{\gamma }$, and stability and interaction properties were not specified. The supplied mass and freeze-out parameter alone therefore do not establish a precise fit.

For odd $N>1$ coprime to $b$, the Fermat pseudoprime test is $b^{N-1}\equiv 1(\mathrm{mod}N)$. The Euler pseudoprime test meant here is the Euler–Jacobi test,where the right side is the Jacobi symbol. Write $N-1=2^{s}d$, with $d$ odd. The strong pseudoprime test passes when $b^d\equiv 1$, or when $b^{2^{j}d}\equiv -1$ for some $0\le j<s$. Terminology often reserves “pseudoprime” for composite numbers; the question uses it for passing the respective test, including primes.

For the final example, $341=11\cdot 31$ is composite and $2^{10}=1024\equiv 1(\mathrm{mod}341)$, so $2^{340}\equiv 1$. It passes the Fermat primality test. However $2^{170}\equiv 1$, whereas the supplementary law for the Jacobi symbol gives $(2/341)=-1$ since $341\equiv 5(\mathrm{mod}8)$. Thus 341 is Fermat but not Euler–Jacobi pseudoprime to base two. A weaker convention defining an Euler test merely by $b^{(N-1)/2}=\pm 1$ would not give the claimed distinction, which is why the Jacobi convention matters.

If $N$ is prime, Fermat's little theorem gives $b^{2^{s}d}\equiv 1$. If $b^d\equiv 1$, the strong test already passes. Otherwise let $j\ge 1$ be the first index with $b^{2^{j}d}\equiv 1$. Its preceding value $u=b^{2^{j-1}d}$ satisfies $u^2=1$ and $u\ne 1$ in the field ${\mathbb{F}}_N$. Since $(u-1)(u+1)=0$ in a field, $u=-1$. The index $j-1$ lies between zero and $s-1$, so every coprime base passes the strong test when $N$ is prime.

For the plain pseudoprime assertion use the Fermat primality test. In the finite group of units modulo an integer, letIt is the kernel of a group homomorphism, because the unit group is an abelian group. The given witness makes $L$ a proper subgroup. By Lagrange's theorem for finite groups, $|L|\le \phi (N)/2$. Thus at least half of all units fail the test. Removing base one removes a passing base, so among the permitted $1<b<N$ there remain $\phi (N)-|L|$ failures out of $\phi (N)-1$ bases, still at least half.

The same argument works for the Euler–Jacobi test: its passing bases form the kernel of $b\mapsto b^{(N-1)/2}(b/N{)}^{-1}$. One must not assert that strong-test passing bases form a subgroup; the kernel argument above concerns the Fermat statement requested here.

The Lie group $PSL(2,\mathbb{C})$ is a second-countable space. For every point of its discrete subgroup, choose the first basic neighbourhood isolating that point; different elements require different such neighbourhoods. This injects $G$ into a countable set, proving countability.

A discrete subgroup is closed and meets compact sets in finitely many points. Indeed, distinct subgroup elements accumulating anywhere would give quotients converging to identity, contradicting an identity neighbourhood containing no other subgroup element. To apply this, use the positive Hermitian determinant-one model of hyperbolic space. At the base point $I$, the image under a determinant-one representative $g$ is $g{g}^{\ast }$ andThis distance formula follows by unitary diagonalization: eigenvalues are $e^t,e^{-t}$ and the axial hyperbolic displacement is $|t|$. Thus bounded displacement bounds all matrix entries. The determinant-one condition is closed, so the resulting set is compact; the quotient by $\{I,-I\}$ is compact too. Conjugation gives the same conclusion at any base point. These are the bounded displacement sets of hyperbolic isometries are compact facts.

Hence, for each $R$, only finitely many $g\in G$ satisfy $d(p,gp)\le R$. Any enumeration of an infinite group without repetitions therefore has $d(p,g_{n}p)\to \infty$. A finite discrete group is a harmless exception to the wording: it has no infinite enumeration of distinct elements, and proper discontinuity is immediate.

For a compact set $C$, put $R={\max }_{x\in C}d(p,x)$. If $gC\cap C\ne \varnothing$, choose $x,gx\in C$. The triangle inequality gives $d(p,gp)\le d(p,gx)+d(x,p)\le 2R$. Only finitely many such $g$ exist, proving properly discontinuous group action. Conversely, if the subgroup is not discrete, there are distinct $g_n\to e$. A closed ball about $p$ then meets its images under infinitely many $g_n$, violating this compact-set definition. Thus discreteness is equivalent to proper discontinuity. Here proper discontinuity allows finite point stabilizers; a stronger convention requiring freeness would exclude discrete elliptic subgroups and is not the equivalence intended.

The Gaussian log likelihood function, apart from a constant, isDifferentiating in $\beta$ gives $X^{\mathsf{T}}(Y-X\hat{\beta })=0$. Full column rank makes $X^{\mathsf{T}}X$ invertible, soThe distribution follows by applying this linear map to the Gaussian error vector. Differentiating the profiled likelihood function in ${\sigma }^2$ givesFor its distribution, $P=X(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}$ is symmetric idempotent, and the residual is $(I-P)ϵ$. The needed form of Cochran's theorem says that quadratic forms in a standard Gaussian vector associated with orthogonal symmetric projection matrices summing to identity are independent random variables with chi-squared distributions with degrees equal to their ranks. Applying it to $P$ and $I-P$ givesThus the maximum-likelihood variance estimate uses divisor $n$ and has expectation $(n-p){\sigma }^2/n$; the unbiased residual estimate would use divisor $n-p$. The fitted coefficient and residual quadratic form are also independent.

Choose $F(z)=(z+1{)}^{1/3}{z}^{1/3}(z-1{)}^{1/3}$, each factor initially defined by polar argument in $(-\pi ,\pi )$. The products of the boundary values agree on $(-\infty ,-1)$, so their apparent cuts there cancel; the product extends analytically to $\mathbb{C}\setminus [-1,1]$ and has $F(z)\sim z$ at infinity. In polar coordinates about the three branch points this choice is $\prod r_j^{1/3}\exp [i\sum {\theta }_j/3]$.

For $0<x<1$, its upper and lower bank values are $e^{\pm i\pi /3}[x(1-x^2){]}^{1/3}$. For $-1<x<0$ they are $e^{\pm 2i\pi /3}[(-x)(1-x^2){]}^{1/3}$. The upper-minus-lower jump on each half is therefore $2i\sin (\pi /3)$ times the positive real integrand. Deforming a positively oriented large circle onto the cut, with the upper bank traversed right to left, givesSmall circles about the branch points contribute zero in the limit. At infinity,so the large-circle integral is $-2\pi i/3$. HenceBoth the branch choice and the orientation are essential to the sign.

Set $t=x^2$, so $dx=dt/(2\sqrt{t})$. ThenThe beta-gamma identity, the Gamma function recurrence and the gamma reflection formula give

A comoving material cell contains the same particles in its perturbed and background descriptions, so its mass is $\bar{\rho }{a}^3{d}^{3}q=\rho d^{3}r$. The Jacobian determinant is $a^3\det (I+{\nabla }_q\psi )$. Expanding the determinant givesSplit the potential into homogeneous and perturbed parts. There is a sign error in the printed hint: ${\nabla }^2{\Phi }_0=4\pi G\bar{\rho }$ requires $\nabla {\Phi }_0=+4\pi G\bar{\rho }r/3$, not the printed negative sign. Its acceleration $-\nabla {\Phi }_0$ agrees with $\ddot{a}/a=-4\pi G\bar{\rho }/3$.

For the perturbation, ${\nabla }_q^2\varphi =4\pi G\bar{\rho }{a}^2\delta =-4\pi G\bar{\rho }{a}^2{\nabla }_q\cdot \psi$. Under the usual homogeneous boundary convention the curl-free force therefore satisfieswhere ${\psi }_L$ is the longitudinal part of the displacement. Expanding the particle acceleration gives $\ddot{r}=\ddot{a}(q+\psi )+2\dot{a}\dot{\psi }+a\ddot{\psi }$. The background acceleration cancels the first term, leavingFor a longitudinal displacement this proves the displayed longitudinal displacement in linear dust perturbations equation with ${\psi }_L=\psi$. For unrestricted displacement, a transverse component has no density contrast and instead obeys ${\ddot{\psi }}_T+2H{\dot{\psi }}_T=0$; the full-vector equation printed in the question thus needs the longitudinal assumption. Taking the divergence removes this qualification and gives

For a flat matter-dominated universe, the Friedmann solution has $a\propto t^{2/3}$, $H=2/(3t)$ and $\bar{\rho }=1/(6\pi G{t}^2)$. Thus the density equation isTrying $t^s$ gives $s(s-1)+4s/3-2/3=0$, with roots $2/3$ and $-1$. ThereforeThe growing mode is proportional to the scale factor. When smooth dark energy or a cosmological constant dominates, the matter source density falls while expansion provides continued damping, suppressing growth. In the late constant-$\Lambda$ limit, $H\to H_{\Lambda }$ and $\bar{\rho }\to 0$, so $\ddot{\delta }+2H_{\Lambda }\dot{\delta }\simeq 0$ and $\delta$ tends to a constant plus a decaying $e^{-2H_{\Lambda }t}$ term. Matter perturbations therefore cease growing appreciably rather than continuing the matter-era law indefinitely.

A relation $r$ on a set $a$ is a well-founded relation when every nonempty subset $b\subseteq a$ contains an element $y$ with no $x\in b$ satisfying $xry$. The well-founded recursion theorem states that a rule assigning a value at $y$ from the already assigned values at its predecessors determines a unique function on $a$; the rule must give a unique set-valued output for every possible predecessor assignment. No choice or transitivity of $r$ is needed.

Apply recursion withThe empty supremum is zero. Well-founded induction makes every value an ordinal; the Axiom schema of replacement collects the values into a set. Choose an ordinal $\alpha$ larger than every value. Then $f:a\to \alpha$, and $xry$ implies $f(x)+1\le f(y)$, hence $f(x)<f(y)$.

Conversely, suppose such an ordinal-valued function exists. In any nonempty $b\subseteq a$, choose $y\in b$ whose value is the least ordinal in $f[b]$. There can be no $x\in b$ with $xry$, because it would have smaller value. Thus $r$ is well-founded, proving the equivalence.

Using the axiom of choice, fix a well-order $\prec$ of $a$. Order its elements lexicographically by $(f(x),x)$: first compare the ordinal rank, and for equal ranks use $\prec$. Every nonempty subset has a smallest rank and then a least element in that rank fibre, so this is a well-order. Since $xry$ strictly increases rank, it respects every original relation pair. Hence every well-founded relation extends to a well-order under choice.

In the usual nondegenerate convention, a strongly regular graph of order $n$ and parameters $(d,a,b)$ is a finite simple $d$-regular graph with $0<d<n-1$ in which adjacent vertices have $a$ common neighbours and distinct nonadjacent vertices have $b$. Its adjacency matrix of a graph $A$ obeysFor the usual nondegenerate parameters with $b>0$ the graph is connected, so eigenvalue $d$ has multiplicity one. On the perpendicular complement of the all-ones vector, $J=0$ and the possible eigenvalues areIf $r$ has multiplicity $m$, then zero trace gives $d+mr+(n-1-m)s=0$. Solving provesThe denominator is positive in this nondegenerate convention. Otherwise all eigenvalues on the all-ones complement would coincide, forcing $A=rI+(d-r)J/n$. All off-diagonal entries would then be the same, making the graph complete or edgeless. If complete graphs are admitted with arbitrary unused $b$, the printed claim needs qualification: $K_2$ with $(d,a,b)=(1,0,2)$ satisfies the common-neighbour conditions vacuously but makes the fraction $0/0$. For complete graphs with nonzero denominator, the spectral calculation still works with an unused root of multiplicity zero.

For the triangle-free graph claim, suppose first there are at least two vertices. Nonadjacent vertices have a common neighbour, so the graph is connected. If $u,v$ are adjacent, put $U=N(u)\setminus \{v\}$ and $V=N(v)\setminus \{u\}$. Every member of $U$ is nonadjacent to $v$, and exactly two of its three common neighbours with $v$ lie in $V$, the third being $u$. By symmetry every member of $V$ has exactly two neighbours in $U$. Counting these edges gives $2|U|=2|V|$, hence $d(u)=d(v)$. Connectivity makes the graph $d$-regular. Counting length-two paths from a vertex to non-neighbours givesHere $a=0$, $b=3$, and the larger-root multiplicity becomes $[n-1+d^2/\sqrt{4d-3}]/2$. Thus $t=\sqrt{4d-3}$ is rational and therefore an integer. It is odd, $d=(t^2+3)/4$, and integrality also gives $t\mid d^2$. Since $16d^2\equiv 9(\mathrm{mod}t)$, one obtains $t\mid 9$, henceThe complete bipartite graph $K_{3,3}$ realizes $d=3$: vertices in the same part have all three vertices of the other part as common neighbours, and vertices in opposite parts are adjacent. There are no triangles.

As literally worded, the one-vertex graph is a vacuous exception: it has no triangles or nonadjacent distinct pairs, but has $d=0$. The displayed positive-degree classification therefore requires at least two vertices. This qualification completes the triangle-free graphs with three common neighbours argument rather than silently excluding a counterexample.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension:

Choose a nonzero vector in an irreducible representation. Its finite group orbit spans a finite-dimensional invariant subspace, which must be the whole representation. The representing matrices of an abelian group commute. Choose an eigenspace of one matrix; all other matrices preserve it. Repeatedly choosing an eigenspace for the restrictions of the remaining matrices gives a common eigenvector. Its line is invariant under the whole group. Irreducibility therefore makes the representation that line, proving every irreducible complex representation of a finite abelian group is one-dimensional.

A faithful representation of dimension one embeds $A$ into ${\mathbb{C}}^{\times }$. Every image element is a $|A|$th root of unity by Lagrange's theorem, so the image is a subgroup of the cyclic group of those roots. It is cyclic, and faithfulness makes $A$ cyclic. Conversely a generator of a cyclic group of order $m$ can be sent to $e^{2\pi i/m}$, giving an injective one-dimensional character. Thus such a faithful representation exists exactly when $A$ is a cyclic group.

For $z\in Z$, the eigenspaces of $\rho (z)$ are invariant under $G$, since every representing matrix commutes with $\rho (z)$. Irreducibility forces one such eigenspace to be all of $W$, so $\rho (z)=\chi (z)I$. The multiplication law gives $\chi (z{z}^{\prime })=\chi (z)\chi (z^{\prime })$, making $\chi$ a one-dimensional, hence irreducible, group representation of $Z$. Therefore