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The Fermat little theorem states that if $p$ is a prime number, then $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$. Equivalently, for $p\nmid a$, $a^{p-1}\equiv 1(\mathrm{mod}p)$.

To prove it when $p\nmid a$, multiplication by $a$ permutes the nonzero residues $1,\dots ,p-1$. Indeed, $ai\equiv aj(\mathrm{mod}p)$ implies $p\mid a(i-j)$, and the Euclid lemma implies $i=j$ in this range. Multiplying all these residues givesNone of the factors of $(p-1)!$ is divisible by $p$, so the factorial has a modular inverse. Cancelling it proves $a^{p-1}\equiv 1$, hence $a^p\equiv a$. When $p\mid a$, the latter identity has both sides zero, completing the proof.

For an odd prime number $p\ne 5$, the numbers $10,p$ are coprime. The Fermat little theorem gives $10^{p-1}\equiv 1(\mathrm{mod}p)$, and consequentlyThese exponents are distinct, so infinitely many are obtained.

Write the repdigit with $n$ copies of $5$ asIf $p\ne 3,5$, choose $n$ to be any positive multiple of $p-1$. Then $9$ has a modular inverse modulo $p$, so $p\mid R_n$. Division by $9$ cannot be used modulo $3$; instead $10\equiv 1(\mathrm{mod}3)$ gives $R_n\equiv 5n(\mathrm{mod}3)$, so $3\mid R_{3j}$ for every $j\ge 1$. If $p=5$, every $R_n$ is divisible by $5$. ThusThe separate treatment of $3$ accounts for the prime factor of the geometric-sum denominator.

For the Fibonacci addition formula, use mathematical induction on $k$, with the assertion applying to every $n\ge 1$. At $k=2$, $F_{n+2}=F_{n+1}+F_n=F_2{F}_{n+1}+F_1{F}_n$. If the formula holds for some $k\ge 2$ and all $n$, apply it at $n+1$ and use the Fibonacci number recurrence:This is the assertion at $k+1$, completing the mathematical induction. ThusTaking $k=n\ge 2$ gives the doubling identity

For the Lucas numbers, $L_2=F_3+F_1=3$, and $L_3=F_4+F_2=4=L_2+L_1$. For $n\ge 4$, both expressions for earlier Lucas numbers are available, andHence $\boxed{L_1=1,L_2=3,L_n=L_{n-1}+L_{n-2}(n\ge 3)}$. The separate $n=3$ check matters because the defining expression for $L_1$ was given separately.

The Euclidean algorithm and the Fibonacci number recurrence give, for $n\ge 2$,Repeating reduces to $\gcd (F_2,F_1)=1$; the case $n=1$ is immediate. Therefore $\boxed{\gcd (F_n,F_{n+1})=1(n\ge 1)}$.

For $n\ge 2$, $L_n=F_n+2F_{n-1}$, so a common divisor $d$ of $F_n,L_n$ divides both $F_n$ and $2F_{n-1}$. Since $F_n,F_{n-1}$ are coprime, Bézout's identity gives integers $s,t$ with $s{F}_n+t{F}_{n-1}=1$. Multiplying by $2$ shows $d\mid 2$. Conversely any common divisor of $F_n$ and $2$ divides $L_n$. Including $n=1$, this proves the stronger greatest common divisor identity

Choose the two inertial frames to have coincident origins at $t=t^{\prime }=0$, and let the origin of $S^{\prime }$ have $x=vt$. The Lorentz transformation isHere ${\gamma }_v$ is the Lorentz factor. Along the particle's trajectory, $dx/dt=u$, soDividing gives the relativistic velocity-addition formulaThe denominator is positive because $|uv|<c^2$. Direct algebra yieldssince both factors in the numerator are positive. Consequently $\boxed{|u^{\prime }|<c}$: a subluminal velocity remains subluminal in every such inertial frame.

With no electric field, the Lorentz force is $F=q\dot{r}\times B\hat{z}$. The equation from Newton's second law and its first integral areThe cyclotron frequency is $|\omega |$; the sign of $\omega$ specifies the sense of rotation. Since $\hat{x}\times \hat{z}=-\hat{y}$, the initial data giveThe component equations are $\dot{x}=\omega y$, $\dot{y}+\omega x=u+a\omega$ and $\dot{z}=v$. For $\omega \ne 0$, differentiating the first givesSolving this linear differential equation and then using $y=\dot{x}/\omega$ givesThe perpendicular projection is a circle of radius $|u/\omega |$ centred at $(a+u/\omega ,0)$, the guiding centre. The velocity parallel to the magnetic field stays constant, producing helical motion in a uniform magnetic field.

When $a\omega +u=0$, this simplifies toThus the trajectory is a helix about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega |$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega >0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.

For $\omega =0$, the Lorentz force vanishes and the solution is instead $r(t)=a\hat{x}+ut\hat{y}+vt\hat{z}$, also the limit of the general formula. The special condition then forces $u=0$.

Let $p$ be the magnitude of the incident particle's momentum. The relativistic energy-momentum relation givesThe stationary particle contributes rest energy $m{c}^2$ and zero momentum. Conservation of four-momentum therefore gives equal photon energies $\epsilon =(E+m{c}^2)/2$. A photon has momentum magnitude $\epsilon /c$, so if the two momentum vectors meet at angle $\theta$,Substitution gives the angular relation for two-photon electron-positron annihilation, whose derivation depends only on the stipulated equal rest masses and equal photon energies:All energies and the angle here are in the laboratory inertial frame.

For perpendicular photon trajectories, $\cos \theta =0$ and hence $E=3m{c}^2$. Since $E=\gamma m{c}^2$ with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For the small-speed limit, set $\beta =v/c\ge 0$ and $\delta =\pi -\theta$. The angle formula impliesUsing $\arcsin z=z+O(z^3)$ therefore gives $\delta =\beta +O({\beta }^3)$, soAt zero incident speed the momenta are equal and opposite, so $\theta =\pi$, consistent with the expansion.

Work in an inertial frame, take all masses to be constant, and assume the internal forces obey the pairwise relation from Newton's third law $F_{ij}=-F_{ji}$. For the angular momentum result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from Newton's second law areSumming them and pairing the internal forces gives

About a fixed point $a$, define the total angular momentum and external torque byDifferentiation produces no velocity term because ${\dot{r}}_i\times m_i{\dot{r}}_i=0$. Grouping internal terms into pairs givesEach final cross product is zero by centrality, proving $\boxed{d{L}_a/dt=G_a}$. Equal and opposite internal forces give momentum conservation when the total external force is zero, but do not in general make their internal torques cancel; the central-force assumption supplies that extra step.

Let $M={\sum }_i{m}_i$ and let the centre of mass be $R=M^{-1}{\sum }_i{m}_i{r}_i$, so $P=M\dot{R}$. For an arbitrary moving reference point $a(t)$, the same differentiation givesTaking $a=R$, the extra term is $-\dot{R}\times M\dot{R}=0$. Therefore the corresponding angular momentum about the centre of mass satisfiesIts definition using velocities relative to $\dot{R}$ agrees with the one above, since ${\sum }_i{m}_i(r_i-R)=0$ makes ${\sum }_i(r_i-R)\times m_i\dot{R}=0$. Thus this result remains true even when the centre of mass accelerates.

Finally take the common mass to be $m$ and $F_i=-k{\dot{r}}_i$, with constant $k$. About either a fixed point or the centre of mass,The angular-momentum equation becomes $\dot{L}=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the angular momentum decay under uniform linear drag isThe common mass and common drag coefficient ensure that the same decay factor applies to every term.

For each $g\in G$, conjugation sends a subgroup $H$ to $gH{g}^{-1}$, a subgroup of the same order and hence still proper. Identity conjugation fixes every subgroup, andThese verify the group action laws.

The stabilizer subgroup of $B$ for this conjugation action is its normalizer $N_G(B)$. Since $B\le N_G(B)$, the orbit-stabilizer theorem gives the number of distinct conjugates asAll these conjugate subgroups contain the identity, so counting that element only once givesThe last inequality uses the properness of $B$, so $[G:B]>1$. There is therefore an element outside this union, which is not conjugate to any element of $B$. This proves that conjugates of a proper subgroup do not cover a finite group, including the case $B=\{1\}$.

By transitivity, choose $h\in G$ with $hy=x$. If $gy=y$, thenThus $hg{h}^{-1}\in B$. This proves the conjugation assertion directly from the transitive group action.

For the complex special linear group in dimension two, the Möbius transformation of a matrix $g$ acts by $z\mapsto (az+b)/(cz+d)$, with $ad-bc=1$. It fixes infinity exactly when $c=0$, so its stabilizer subgroup isWe distinguish the cases to avoid dividing by a vanishing coefficient.

If $c\ne 0$, infinity is not fixed, and the finite fixed points solveNeither root is the pole $-d/c$: substituting that value gives $(ad-bc)/c=1/c\ne 0$. The fixed-point set is thereforewith the two entries coinciding when the discriminant is zero. The discriminant identity uses $(a-d{)}^2+4bc=(a+d{)}^2-4$.

If $c=0$ and $a\ne d$, the fixed points are $\{\infty ,b/(d-a)\}$. If $c=0$, $a=d$ and $b\ne 0$, only infinity is fixed. If $c=0$, $a=d$ and $b=0$, the determinant condition gives $g=\pm I$ and every point is fixed. Both central matrices induce the identity Möbius transformation.

Every case has a fixed point, because a quadratic over the complex numbers has a root. The action is transitive: any finite $z_0$ is sent to infinity byand infinity already needs no change. Applying the first argument to a fixed point proves every element of $S{L}_2(\mathbb{C})$ is conjugate to an element of $B$.