First apply the Cantor theorem by diagonalization. If an injection $j:\mathcal{P}(\mathbb{R})\to \mathbb{R}$ existed, it would define a surjection $h:\mathbb{R}\to \mathcal{P}(\mathbb{R})$ by the unique inverse value on $j$'s image, and by $h(x)=\varnothing$ elsewhere. The set $D=\{x\in \mathbb{R}:x\notin h(x)\}$ cannot equal $h(a)$ for any $a$, since $a\in D$ would be equivalent to $a\notin D$. This contradicts surjectivity. There is no injection from the power set of the real line into the real line.

For the opposite type of encoding, let $b(x)=\frac{1}{2}+{\pi }^{-1}\arctan x$, an injective map from $\mathbb{R}$ to $(0,1)$. Give $b(x)$ its canonical binary expansion, choosing the terminating expansion with trailing zeros whenever two expansions exist. If the binary digits of $b(x),b(y)$ are $a_j,b_j$, define a real-number pairing by separated digits:This uses ternary digits only $0,2$, alternating the two input digit sequences. If two outputs first differ at ternary position $k$, that difference has magnitude $2\cdot 3^{-k}$, while every later digit together contributes at most ${\sum }_{j>k}2\cdot 3^{-j}=3^{-k}$. Their outputs therefore differ. The digits recover both binary sequences, hence both inputs. Neither output endpoint $0$ nor $1$ occurs because the input numbers lie strictly between zero and one. Thus $H$ is an injection from ${\mathbb{R}}^2$ into $(0,1)\subset \mathbb{R}$. The ternary construction avoids an unjustified decimal-interleaving argument at ambiguous expansions.

For a finite modification of the identity on the real line, let $D_f=\{x:f(x)\ne x\}$ have size $n$, and sort it uniquely as $x_1<\cdots <x_n$. Its finite record is $(x_1,f(x_1),\dots ,x_n,f(x_n))$. It determines $f$ completely, with the identity used outside the recorded points. Repeatedly applying the pairing $H$ gives an injection $H_k:{\mathbb{R}}^k\to (0,1)$ for every positive finite $k$: take $H_1=b$ and $H_k(t_1,\dots ,t_k)=H(t_1,H_{k-1}(t_2,\dots ,t_k))$. Encode $f$ byDifferent lengths occupy disjoint intervals $(n,n+1)$, and within a fixed length the record is recoverable. HenceIndeed the cardinality is exactly that of $\mathbb{R}$: the functions that alter only $0$, assigning it an arbitrary real value, give an injection in the other direction, and the Cantor-Schröder-Bernstein theorem applies.

Fermat's little theorem says that for a prime $p$ and $p\nmid a$, $a^{p-1}\equiv 1(\mathrm{mod}p)$; equivalently $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$. The Wilson theorem says $(p-1)!\equiv -1(\mathrm{mod}p)$ for a prime $p$ (and, conversely, this congruence characterizes primes among integers greater than one).

For $p=2$, $x=1$ is a square root of minus one modulo a prime. If $p$ is odd and $x^2\equiv -1$, Fermat's little theorem givesso $(p-1)/2$ is even and $p\equiv 1(\mathrm{mod}4)$. Conversely, put $h=(p-1)/2$. Pairing $j$ with $p-j$ in the factorial gives $(p-1)!\equiv (-1{)}^h(h!{)}^2$. For $p\equiv 1(\mathrm{mod}4)$, $h$ is even, so the Wilson theorem yields $(h!{)}^2\equiv -1$. Thusand in the latter case $x=h!$ supplies a solution.

For the multiplicative order assertion, divide $k$ by $d$: $k=qd+r$, $0\le r<d$. Since $x^d\equiv 1$ and $x^k\equiv 1$, it follows that $x^r\equiv 1$. Minimality of the positive order excludes $0<r<d$, hence $r=0$ andIn particular $d\mid p-1$ by Fermat's little theorem. Negative $k$, if included, are handled by the same division using the modular inverse of $x$.

A Fermat number in the paper has $n\ge 1$. If a prime $p$ divides $F_n=2^{2^n}+1$, it is odd and $2^{2^n}\equiv -1(\mathrm{mod}p)$. Squaring gives $2^{2^{n+1}}\equiv 1$, so the order $d$ divides $2^{n+1}$. It does not divide $2^n$, since $-1\ne 1$ modulo an odd prime. Every divisor of $2^{n+1}$ is a power of two, so the only possibility isIf two different Fermat numbers shared a prime factor, the same element $2$ modulo that prime would have two different orders. This is impossible, proving pairwise coprimality. Also $2^{n+1}\mid p-1$, so for $n\ge 1$ every such prime is $1$ modulo $4$. No prime $3$ modulo $4$ can occur.

A prime $1$ modulo $4$ need not occur: take $p=13$. It is prime, $2^6=64\equiv -1(\mathrm{mod}13)$, and $2^4\equiv 3\ne 1$. Since the order divides $12$, the first congruence excludes every divisor of $6$ and the second excludes $4$; hence the order is $12$. This is not a power of two, so $13$ is not a prime divisor of a Fermat number. The index restriction matters: the conventional extra number $F_0=3$ is outside the paper's positive-$n$ family.

For $0<v<c$, the Lorentz transformation with coincident origins isFor two simultaneous events in $S$, $\Delta t=0$, so $\Delta t^{\prime }=-\gamma v\Delta x/c^2$. Distinct spatial positions therefore generally give different times in the boosted frame: this is relativity of simultaneity.

Choose the emission events at $(t,x)=(0,0)$ and $(0,d)$. After emission, the photon world lines are $x=ct$ and $x=d+ct$. Applying the Lorentz transformation to either givesThus the two photons move at speed $c$ in $S^{\prime }$ on parallel lines whose intercepts differ by $\gamma (1+v/c)d$. Measuring their positions at the same $t^{\prime }$ after both emissions gives the photon separation under a collinear Lorentz boost:This separation is constant. It is not the contracted distance between the stationary sources: the two emission events are not simultaneous in $S^{\prime }$, and one photon has already moved when the other is emitted.

By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe centre moves on a straight line with constant velocity; zero velocity is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

Yes: the two equal masses can follow the same fixed circle in diametrically opposite positions. Put the stationary centre of mass at the circle's center and choose tangential velocities with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational force is directed toward that center, with magnitude $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.