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By orthogonal diagonalization of the positive-definite symmetric matrix $A$, form its positive symmetric inverse square root $S=A^{-1/2}$. Then $S^{T}AS=I$, and $C=SBS$ is real symmetric. Let $Q$ have an orthonormal eigenbasis of $C$ as its columns, so $Q^{T}CQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Taking $P=SQ$ givesBoth factors of $P$ are invertible. This proves the simultaneous congruence diagonalization with a positive form by an explicit construction; it is congruence rather than simultaneous similarity.

To find the entries without constructing $P$, take determinants inSince $\det P\ne 0$, the entries are precisely the roots, with multiplicity, of $\det (B-\lambda A)=0$. Equivalently they are the eigenvalues of $A^{-1}B$, although that matrix need not itself be symmetric.

For the given $A$, its leading principal minors are $2,3,4$, so it is positive definite. Expanding the determinant givesThus the diagonal entries, in any order, are

For an inviscid fluid of constant density, with irrotational velocity $\mathbf{u}=\nabla \Phi$ and conservative body-force potential $\Psi$, the Unsteady Bernoulli equation isIt holds throughout a connected irrotational fluid region, up to the time-dependent gauge of $\Phi$. Indeed the Euler equations and $\nabla \times \mathbf{u}=0$ imply that the gradient of the left side is zero. In the bubble problem there is no body-force contribution compatible with the assumed spherical symmetry.

Spherical incompressibility gives ${\partial }_r(r^2{u}_r)=0$. The moving interface imposes $u_r(R,t)=\dot{R}$, soThis potential tends to zero at infinity, where the velocity tends to zero and $C(t)=p_{\infty }/\rho$. Differentiate $\Phi$ at fixed spatial $r$, not along the moving interface: ${\Phi }_t=-(2R{\dot{R}}^2+R^2\ddot{R})/r$. At $r=R$, pressure continuity without surface tension gives $p=p_{\infty }-\Delta p$. Bernoulli therefore yields $-2{\dot{R}}^2-R\ddot{R}+{\dot{R}}^2/2-\Delta p/\rho =0$, or the Rayleigh equation for an inviscid spherical bubble

Integrating the kinetic energy over the exterior fluid givesDifferentiate and use the radius equation:With $R(0)=R_0$ and $\dot{R}(0)=0$, the pressure-work energy balance for a spherical bubble integrates toorThe collapsing branch takes the negative square root for $\dot{R}$. The energy gain is exactly the pressure difference times the lost bubble volume, providing a check on the sign and numerical factor.

For $g(z)=(az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, direct subtraction from its conjugate givesThe imaginary part remains positive, and the inverse transformation has the same form, so $g$ is a bijection of the upper half-plane. MoreoverThus every path has the same hyperbolic length as its image. Taking the infimum over connecting paths, and using the inverse for the reverse inequality, proves every $g\in \Gamma$ is an isometry.

LikewiseIts squared modulus and the two imaginary-part factors cancel in the quotient, proving

We can normalize any ordered distinct pair to $(i,iy)$ with $y>1$. First send $z=x_0+i{y}_0$ to $i$ by $(\zeta -x_0)/y_0$, a real determinant-one Möbius transformation after rescaling its representing matrix. The stabilizer transformationsfix $i$. Under the Cayley map $C(\zeta )=(\zeta -i)/(\zeta +i)$, direct substitution gives $C(k_{\tau }\zeta )=e^{2i\tau }C(\zeta )$. Choose $\tau$ to rotate the second point's nonzero disc coordinate onto the positive real axis. Its inverse image is $iy$ with $y=(1+|C|)/(1-|C|)>1$.

The vertical path from $i$ to $iy$ has length $\log y$. Any other path has length at least $\int |dy|/y\ge |\int dy/y|=\log y$, so $\rho (i,iy)=\log y$. ConsequentlyInvariance of both quantities transfers this equality back to every pair; coincident points satisfy it as well. Therefore

For the right-triangle identity, normalize its right vertex to $i$ and one leg to the vertical geodesic, with endpoint $iy$, $y>1$. The other leg lies on the unit semicircle, with endpoint $s+it$, $s^2+t^2=1$, $t>0$. These geodesics meet orthogonally at $i$, since the metric is conformal to the Euclidean metric. The semicircle is a geodesic because a real Möbius transformation sends it to a vertical line; vertical lines are minimizing by the same length bound. The stabilizer rotations above can align any given tangent direction, giving this normal form for every right triangle.

If the legs have lengths $a,b$ and the opposite side has length $c$, the distance formula givesandThus the hyperbolic Pythagorean theorem is

Treat the two row totals as fixed sample sizes, with independent multinomial observations within each row. More generally let $N_{ij}$ be an $r$ by $c$ table, row totals $n_i$, column totals $m_j$ and total $N$. Under the unrestricted alternative, row $i$ has its own category probabilities $p_{ij}$; independence means that all rows have the same probabilities $p_j$.

Ignoring the multinomial coefficients, which cancel in a likelihood ratio, the unrestricted likelihood is ${\prod }_{ij}{p}_{ij}^{N_{ij}}$. Maximizing ${\sum }_j{N}_{ij}\log p_{ij}$ under ${\sum }_j{p}_{ij}=1$ gives ${\hat{p}}_{ij}=N_{ij}/n_i$. Under the null, the log likelihood is ${\sum }_j{m}_j\log p_j$, giving ${\hat{p}}_j=m_j/N$. Thus the null fitted counts are $E_{ij}=n_i{m}_j/N$, and the generalized likelihood ratio isA zero observed count contributes zero by continuity. Small $\Lambda$, equivalently large $G^2$, is evidence against independence. The unrestricted model has $r(c-1)$ free probabilities and the null has $c-1$, a difference of $(r-1)(c-1)$. Under the usual large-sample likelihood-ratio approximation at an interior null, $G^2$ has a chi-squared distribution with that many degrees of freedom. This is the likelihood-ratio test of independence in a contingency table. Sampling the entire table multinomially instead gives the same statistic and degrees of freedom after estimating the row probabilities.

For the supplied data the fitted counts areSubstitution givesThere are $(2-1)(3-1)=2$ degrees of freedom. The supplied 1% critical value is $9.21$, and $0.9701<9.21$. Do not reject independence at the 0.01 level. The approximate chi-squared tail probability is $e^{-G^2/2}\simeq 0.616$. This lack of rejection does not establish that the classifications are independent; the test supplies no evidence against that null here. All fitted counts are at least ten, making the stated asymptotic approximation reasonable.

Put $\mu =\cos \vartheta$ and separate $\Phi =R(r)Y(\mu )$. Multiplying Laplace's equation by $r^2/(RY)$ givesFor the regular polynomial angular modes, $\lambda =n(n+1)$ and $Y=P_n(\mu )$ is a Legendre polynomial. The radial equation has branches $r^n$ and $r^{-n-1}$. Regularity at the origin keeps only the first inside, and decay at infinity keeps only the second outside. Continuity at $r=1$ identifies their coefficients, so a mode hasIts outward-minus-inward radial derivative is $-(2n+1)a_n{P}_n(\mu )$. The prescribed angular dependence decomposes as $V{\mu }^2=V{P}_0/3+2V{P}_2/3$, since $P_0=1$ and $P_2=(3{\mu }^2-1)/2$. Hence $a_0=-V/3$ and $a_2=-2V/15$, with no other required modes. The spherical harmonic matching with a derivative jump givesBoth branches are harmonic in their respective regions. They agree at the sphere, and their derivative jump is $V/3+(2V/3)P_2=V{\cos }^2\vartheta$. The interior is regular at zero and the exterior decays at infinity. An additional homogeneous matching mode would satisfy $-(2n+1)a_n=0$, so its coefficient vanishes.

Take any sequence $(y_n)$ in $Y$. Surjectivity lets us choose $x_n\in X$ with $f(x_n)=y_n$. By sequential compactness of $X$, some $x_{n_j}\to x\in X$. Continuity gives $y_{n_j}=f(x_{n_j})\to f(x)\in Y$. Thus $Y$ has the Bolzano–Weierstrass property.

Now suppose $f$ is a bijection. To prove continuity of its inverse, let $y_n\to y$ and put $x_n=f^{-1}(y_n)$, $x=f^{-1}(y)$. If $x_n\nrightarrow x$, there are $\epsilon >0$ and a subsequence with $d(x_{n_j},x)\ge \epsilon$. Sequential compactness supplies a further subsequence converging to $x^{\prime }$. By continuity of $f$, $f(x^{\prime })=\lim y_{n_j}=y=f(x)$. Injectivity implies $x^{\prime }=x$, contradicting the fixed lower distance bound. Hence $x_n\to x$. Sequential continuity is equivalent to continuity in metric spaces, proving $f^{-1}$ is continuous.

Apply this to $g(t)=e^{it}$ on $[-\pi /2,\pi /2]$, mapping onto the closed right semicircle $S$. This interval is sequentially compact, and $g$ is a continuous bijection: two equal exponential values differ by an integral multiple of $2\pi$, impossible for distinct points of an interval of length $\pi$. Consequently $g^{-1}:S\to [-\pi /2,\pi /2]$ is continuous. For $\mathrm{Re}z>0$, the normalization $z\mapsto z/|z|$ is continuous and takes values in the open right semicircle. Thereforeis a continuous argument satisfying $z=|z|e^{i\mathrm{Arg}z}$. Equivalently it is $\arctan (\mathrm{Im}z/\mathrm{Re}z)$.

For distinct positions and positive masses, Newton's law of universal gravitation givesThe proposed homothetic gravitational motion has ${\ddot{\mathbf{x}}}_i=-(2/9)t^{-4/3}{\mathbf{a}}_i$ and pair distances $t^{2/3}|{\mathbf{a}}_i-{\mathbf{a}}_j|$, so cancellation of $t^{-4/3}$ gives the time-independent equationsThese define a Newtonian central configuration with this choice of scale. Fordifferentiation in ${\mathbf{a}}_i$ gives ${\nabla }_{{\mathbf{a}}_i}\Phi =(2/9)m_i{\mathbf{a}}_i-G{\sum }_{j\ne i}{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3=0$, exactly the displayed equations. The double sum in the paper counts each pair twice, so it agrees with this expression.

Multiply the configuration equations by $m_i$ and sum. Pair contributions cancel, giving ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Thus the total linear momentum is $\mathbf{P}=(2/3)t^{-1/3}{\sum }_i{m}_i{\mathbf{a}}_i=0$. Each position is parallel to its velocity, so every $m_i{\mathbf{x}}_i\times {\dot{\mathbf{x}}}_i=0$ and the total angular momentum is zero. About any other fixed point $\mathbf{b}$, it is ${\mathbf{L}}_{\mathbf{b}}={\mathbf{L}}_O-\mathbf{b}\times \mathbf{P}=\boxed{0}$.

For $b>1$, Euler's theorem allows $d=\phi (b)>0$, since $a,b$ are coprime integers. At $b=1$ every congruence holds and one may take $d=1$. Let $y$ be the multiplicative order of $a$ modulo $b$. Write a nonnegative exponent $x=qy+r$ with $0\le r<y$ using the Euclidean division. Then $a^x\equiv a^r(\mathrm{mod}b)$, so $a^x\equiv 1$ implies $r=0$ by minimality of $y$. Thus $y\mid x$. Negative exponents, if included, give the same conclusion using the inverse residue of $a$.

A prime divisor of a Fermat number $F_n=2^{2^n}+1$ is odd, and satisfies $2^{2^n}\equiv -1(\mathrm{mod}p)$. Therefore $2^{2^{n+1}}\equiv 1(\mathrm{mod}p)$, so its multiplicative order divides $2^{n+1}$. It does not divide $2^n$, since $-1\ne 1$ modulo the odd prime. Every divisor of $2^{n+1}$ is a power of two, henceBy Fermat's little theorem and the divisibility just proved, $2^{n+1}\mid p-1$, giving $\boxed{p\equiv 1(\mathrm{mod}{2}^{n+1})}$.