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If $x$ is a fixed point of $f$, thenConversely, applying the same argument to $f={\sigma }^{-1}g\sigma$ transports every fixed point of $g$ back to one of $f$. Thus the bijection $\sigma$ restricts to a bijection between the two fixed-point sets, so conjugate functions have the same number of fixed points.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$; equivalently, its elements can be listed in a sequence. The integers are countable, for example in the order $0,1,-1,2,-2,\dots$. Enumerating pairs by diagonals, or applying the Cantor pairing function, then shows that $\mathbb{N}\times \mathbb{Z}$ is countable.

Suppose instead that the power set $\mathcal{P}(\mathbb{N})$ had an enumeration $A_1,A_2,\dots$. The Cantor diagonal argument formsFor every $n$, the set $D$ differs from $A_n$ at the element $n$, so it is absent from the alleged enumeration. Therefore

In the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, and choose $e$ coprime toThe decryption exponent is the modular multiplicative inverseThus $ed=1+k\varphi (N)$. For a message with $\gcd (m,N)=1$, the Fermat-Euler theorem givesFor $N=187=11\cdot 17$, $\varphi (N)=160$. The extended Euclidean algorithm givesso

The Fermat-Euler theorem states that if $\gcd (a,n)=1$, thenLet $r_1,\dots ,r_{\varphi (n)}$ be a reduced residue system modulo $n$. Multiplication by $a$ permutes these classes, because $a{r}_i\equiv a{r}_j(\mathrm{mod}n)$ implies $r_i\equiv r_j(\mathrm{mod}n)$. HenceEvery $r_i$ is invertible modulo $n$, so cancellation gives $a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$.

The Euler totient function $\varphi (n)$ is the number of residue classes modulo $n$ that are coprime to $n$. The Chinese remainder theorem says that for pairwise coprime moduli $m_i$, reduction induces a bijectionIt restricts to a bijection of unit groups. If $n=p_1\cdots p_r$ is squarefree, thenEach of the primes $3,5,17,257$ is one more than a power of two, so every squarefree product of them has power-of-two totient. Ten odd examples are

For the rotation group $C_5$, the identity fixes all $n^5$ colorings. Because five is prime, each of the four nonidentity rotations is a single five-cycle and fixes exactly the $n$ constant colorings. Burnside lemma therefore gives the number of coloring necklaces

For the dihedral group $D_{10}$, the five reflections each fix one vertex and interchange two pairs of vertices, so each fixes $n^3$ colorings. Adding these to the identity and rotation contributions gives the number of coloring bracelets

The singleton orbits are exactly $X^G$. Every other orbit represented by $x$ has size $|G|/|G_x|$ by the orbit-stabilizer theorem. Partitioning $X$ into its orbits of a group action therefore gives

For the second identity, countin two ways. Holding $g$ fixed givesHolding $x$ fixed gives $|Z|={\sum }_{x\in X}|G_x|$. Each orbit contributes $|G|/|G_x|$ points, each with stabilizer size $|G_x|$, and hence contributes $|G|$. Thus $|Z|=|G||X/G|$, proving Burnside lemma:

False. Compositions of scalings $z\mapsto \mu z$ and inversion $z\mapsto 1/z$ reduce to one of the forms $\nu z$ or $\nu /z$; they preserve the unordered pair $\{0,\infty \}$. Butis a Möbius transformation fixing $0$ and sends infinity to $1$, so it cannot be such a composition.

True. A Möbius transformation fixes infinity exactly when $c=0$, so it has the affine formThis is the composition of the scaling $z\mapsto \alpha z$ and the translation $z\mapsto z+\beta$.

True. Conjugate the unique fixed point to infinity. The resulting map lies in the Affine subgroup of the Möbius group, so it has the form $g(z)=az+b$. A unique fixed point forces $a=1$ and $b\ne 0$; otherwise $g$ would have a finite fixed point as well. Conjugation by the scaling $z\mapsto z/b$ changes $z\mapsto z+b$ into $z\mapsto z+1$. Thus a nonidentity Möbius transformation with one fixed point is conjugate to a translation.

Write the Möbius transformation asA finite fixed point of a Möbius transformation satisfiesIf three distinct points of the Riemann sphere are fixed, this quadratic is identically zero, with the usual interpretation when one point is infinity. Hence $c=b=0$ and $d=a$, so $f(z)=z$. A nonidentity transformation's fixed-point equation is a nonzero polynomial of degree at most two on the sphere, and therefore it has one repeated fixed point or two distinct fixed points.

Now $E=u^2/2-k/R<0$, so the orbit is bound. Since $L=ub$, the initial energy isAt the farthest central-force radial turning point $q$, the radial speed vanishes and the transverse speed is $L/q$, givingEquating these expressions and multiplying by $2q^2$ yieldsThus $A=L^2/b^2-2k/R$, $B=2k$, and $C=-L^2$.

The central force conserves angular momentum. The initial impact parameter gives $L=ub$, while at closest approach the velocity is transverse and $L=wp$. Thus $u=wp/b$. Conservation of energy givesUnder $u^2\gg 2k/R$, neglect the initial potential term and substitute $u=wp/b$:Consequently

Green theorem states that for a positively oriented planar region $D$,Here $P=x{y}^2-\lambda y$ and $Q=x^{2}y$, soThe stated orientations make $C_1\cup C_2$ the positive boundary of the region inside the radius-$a$ circle and outside the radius-$b$ circle. Its area is $\pi (a^2-b^2)$ regardless of the inner circle's center. Therefore

Set $F(x)=|x^3-x|$. The relation is exactlyIt is therefore the equivalence relation induced by a function: equality of the two function values is reflexive, symmetric, and transitive. Hence this is an equivalence relation.

This relation is reflexive because choosing $n=1$ gives $|x/x|=1$. It need not be symmetric: take positive real numbers $y=e$ and $x=e^2$. Then $|x/y^2|=1$ is rational, so $x\sim y$. Butis transcendental, hence irrational, for every integer $m\ge 1$. Thus $y\nsim x$, and the relation is not an equivalence relation.