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There is a stronger conclusion than the requested limit. For $x>0$, the event $N_n-n\ge x\sqrt{n}$ forces $N_n\ge n+1$, because $N_n$ is an integer. In particular, at least $n$ arrivals have occurred by time $n$, so the $n$-th Poisson arrival time satisfies $T_n\le n$. This contradicts $T_n-n\ge y\sqrt{n}$ for $y>0$. Thus the two events are disjoint:

Let $X_f$ have the specified probability density function $f_m$. Splitting the integral into the positive and negative half-lines, and using the Gamma integral ${\int }_0^{\infty }{x}^r{e}^{-sx}dx=r!/s^{r+1}$ for a nonnegative integer $r$ and $s>0$, givesAt $t=2$ or $t=-2$, one term becomes the divergent integral of $x^m$; beyond these endpoints it grows exponentially. Consequently, the full extended-valued moment-generating function isIn particular $M_f(0)=1$, checking normalization of a probability distribution. All absolute moments exist. Direct integration yieldsThe symmetric distribution makes every odd moment zero, so

Now let $X_g$ have probability density function $g_m$. Its moment-generating function would beFor every $t\ne 0$, one summand has exponential growth, which dominates the polynomial decay. HenceThere is no finite moment-generating function on any neighborhood of zero, but some moments still exist. Specifically,ThusFor even $k\ge m+1$, the moment is $+\infty$, so it is not a finite moment. For odd $k\ge m+1$, both the positive and negative contributions have infinite integral, so the expected value is undefined. Cancellation in a Cauchy principal value does not define an expected value. In particular, when $m=0$ there are no finite positive-order moments.

A return path with $u$ upward steps and $v$ downward steps satisfies $au=bv$. Since $a,b$ are coprime, this is equivalent to $u=bk$ and $v=ak$ for an integer $k\ge 1$. Its length is therefore $k(a+b)$.

Conversely, for each such $k$, any ordering of $bk$ upward and $ak$ downward steps is an allowed positive-probability return path. The possible strictly positive return times are exactly the positive multiples of $a+b$. By the definition of the period of a state in a Markov chain as their greatest common divisor,In particular, none of the states is aperiodic.

Fix any two integers $i,j$. By Bézout's identity, there are integers $u,v$ with $au-bv=j-i$. Replacing $(u,v)$ by $(u+kb,v+ka)$ leaves this identity unchanged. Choose $k$ large enough that both entries are nonnegative. A path with $u+kb$ upward steps and $v+ka$ downward steps then takes $i$ to $j$, with strictly positive probability $p^{u+kb}{q}^{v+ka}$. There are no boundary restrictions on this state space.

This applies in both directions, so every pair of states communicates. Hence

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

Take an integer $c\ge \max \{a,b\}$. On $I=\{-c,\dots ,c\}$, the modified transition matrix $Q=(q_{i,j})$ agrees with $P$. Above $c$, it stays put with probability $a/(a+b)$ and jumps to each of $c-b+1,\dots ,c$ with probability $1/(a+b)$. Below $-c$, it stays put with probability $b/(a+b)$ and jumps to each of $-c,\dots ,-c+a-1$ with probability $1/(a+b)$. All unspecified entries are zero. These rows sum to one. The integer qualification is necessary: if $c$ were nonintegral, $i=\lfloor c\rfloor +1$ would fit none of the printed row conditions.

DefineAn interior jump cannot go below $-c-b$ or above $c+a$, and every exterior reset lands in $I$. Thus $C$ is closed.

To prove that it is one communicating class, start anywhere in $C$. From $L$ or $U$, a reset can reach $I$ with positive probability. From any state of $I$, repeated $+a$ steps reach some state of $U$, after staying in $I$ up to the last step. A reset from this state can reach every point of the consecutive blockEvery integer $j\in \{-c-b,\dots ,c\}$ has the form $t-kb$ for some $t\in B$ and an integer $k\ge 0$. Repeated $-b$ steps from $t$ reach $j$, with all departure states in $I$ until a possible last step into $L$. This gives access to all of $I\cup L$. Finally, each $j\in U$ can be reached from $j-a\in I$. Every path described has strictly positive probability. Therefore every state in $C$ reaches every other state in $C$.

Each state outside $C$ can only stay put or reset into $I$, and no transition from $C$ returns to it. Different states outside $C$ cannot reach one another. Hence the complete list of communicating classes isThe finite class $C$ is the unique closed communicating class. The exterior singleton classes are transient communicating classes, since their self-transition probabilities are strictly less than one. Notice that this reachability proof does not require $a,b$ to be coprime.

For a ring $R$, its ring characteristic is the least positive integer $n$ with $n{1}_R=0$, or zero if no positive such integer exists. Equivalently it is the least positive integer satisfying $nr=0$ for every $r\in R$, because $nr=(n{1}_R)r$. This additive formulation also defines the characteristic of a rng when an identity is not assumed.

Let $R$ be an integral domain. Since $1_R\ne 0$, its characteristic is not one. If its positive characteristic were a composite integer $n=ab$ with $1<a,b<n$, minimality would give $a{1}_R\ne 0$ and $b{1}_R\ne 0$, butThis contradicts the absence of zero divisors. ThereforeThis is the characteristic of an integral domain theorem. If an integral domain is defined without requiring an identity, the same conclusion follows from the additive definition: for a hypothetical composite characteristic $n=ab$, choose $r$ with $ar\ne 0$, which exists by minimality of $n$. Then $(ar)(bs)=n(rs)=0$ for every $s$, forcing $bs=0$ for every $s$ and contradicting minimality again.

Since ${\omega }^2=-1-\omega$, products of integer linear combinations of $1,\omega$ remain of the same form. Thus the Eisenstein integers $E=\mathbb{Z}[\omega ]$ form a subring of $\mathbb{C}$, and hence an integral domain. Their Eisenstein norm isIt is a positive integer on nonzero elements and is multiplicative, because the complex modulus is multiplicative.

For $\alpha ,\beta \in E$ with $\beta \ne 0$, write $\alpha /\beta =u+v\omega$ with real $u,v$. Choose integers $m,n$ with $|u-m|,|v-n|\le 1/2$, and let $q=m+n\omega$. With $s=u-m$ and $t=v-n$,Consequently $r=\alpha -q\beta$ belongs to $E$ and satisfies $N(r)\le \frac{3}{4}N(\beta )<N(\beta )$ when $r\ne 0$. This proves $E$ is a Euclidean domain with Euclidean function $N$, the Euclidean norm on the Eisenstein integers.

For the representation of the prime $p$, first note that $3=N(1-\omega )$. Now suppose $p>3$ and $p\mid n^2+3$. The factorizationshows that the Eisenstein integer $p$ divides a product. It divides neither factor: an element $a+b\omega$ lies in $pE$ exactly when both integer coefficients are divisible by $p$, and the second coefficients here are $2$ and $-2$. Hence $p$ is not a prime element of $E$.

To justify the necessary factorization step, in any principal ideal domain an irreducible element $\pi$ is prime. If $\pi \nmid a$, write $(\pi ,a)=(d)$. Since $d\mid \pi$, irreducibility makes $d$ a unit or an associate of $\pi$. The latter would give $\pi \mid a$, so $d$ is a unit and $u\pi +va=1$ for some $u,v$. If $\pi \mid ab$, multiplication of this identity by $b$ gives $\pi \mid b$. This proves irreducible elements of a principal ideal domain are prime without assuming unique factorization.

By part (a), $E$ is a principal ideal domain. The integer $p$ has norm $p^2>1$, so it is a nonzero nonunit. Since it is not prime, the result just proved implies that it is reducible: $p=uv$ with $u,v$ both nonunits. A nonzero element of $E$ is a unit exactly when its norm is one, since $N(z)=1$ gives inverse $\overline{z}$ and multiplicativity proves the converse. ThusThe only possible positive integer factors exceeding one are $N(u)=N(v)=p$. Writing $u=a+b\omega$ givesThis is the prime representation by the Eisenstein norm argument.

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

For every $i\in {\mathbb{N}}_0$, truncate $w$ before coordinate $i$:This $f_i$ lies in $F$. For $j<i$, $(w-f_i)(j)=0$. For $j\ge i$, $(w-f_i)(j)=w(j)=\pm 2^j$, which is divisible by $2^i$. In particular the integer sequencebelongs to $B$ and satisfies $w=f_i+2^i{g}_i$.

Apply the module homomorphism $\pi$. Since $f_i\in F\subseteq \ker \pi$,Fixing any coordinate $j$, the integer $\pi (w)(j)$ is therefore divisible by $2^i$ for every $i$. The only such integer is zero: if $a\ne 0$, choose $i$ with $2^i>|a|$, which rules out $a=2^{i}b$ for any nonzero integer $b$. Thus every coordinate of $\pi (w)$ vanishes, contradicting part (b)(iii).

It follows thatThe decisive property is vanishing of infinitely divisible integer sequences, namely ${\bigcap }_{i\ge 0}{2}^{i}B=\{0\}$.

The Gaussian integers form an integral domain, since they are a subring of the complex numbers. Use the Gaussian integer normwhich is a positive integer for nonzero $a+bi$ and satisfies $N(\alpha \beta )=N(\alpha )N(\beta )$.

For $\alpha ,\beta \in \mathbb{Z}[i]$ with $\beta \ne 0$, write $\alpha /\beta =x+iy$. Choose integers $m,n$ with $|x-m|\le 1/2$ and $|y-n|\le 1/2$, put $q=m+in$, and set $r=\alpha -\beta q$. ThenThus division with remainder has either $r=0$ or a strictly smaller Gaussian integer norm. This proves that $N$ is a euclidean function and

Each entry of the matrix $Q$ from part (b)(iii) has unique real and imaginary integer coefficients. WriteIt already belongs to the general linear group ${\mathrm{GL}}_2(\mathbb{Z}[i])$. Expanding the relation from part (b)(iii) givesThe real and imaginary parts must both vanish, soThese are the required identities, with $e_2=(0,1{)}^{\mathsf{T}}$.

Euler's criterion states, for an odd prime $p$,Apply it to $a=-1$. The integer $(-1{)}^{(p-1)/2}$ is $1$ when $p\equiv 1(\mathrm{mod}4)$ and $-1$ when $p\equiv 3(\mathrm{mod}4)$. Since those signs are distinct modulo an odd prime, the modular congruence determines the actual Legendre symbol:This is the first supplementary law for quadratic reciprocity.

For an odd prime $p$, the Legendre symbol isIt depends only on the congruence class of the integer $a$. The values $1$ and $-1$ distinguish a nonzero quadratic residue from a quadratic nonresidue; zero is a separate case.

Choose any pair $(a,b)$ from part (i), and put $A=2^a{3}^b$. The divisibilities of the exponents makepositive integers. Their powers areThus $x^3+y^5=z^7$, with all three powers nonzero and distinct. For example, take $a=84+105t$, $b=90+105t$ for every nonnegative integer $t$. Then $A$ strictly increases, giving infinitely many distinct solutions. The construction uses modular congruences to turn the elementary identity $A+2A=3A$ into three prescribed perfect powers.

Writing $a=21r$, the second condition becomes $r\equiv 4(\mathrm{mod}5)$. Writing $b=15s$, its second condition becomes $s\equiv 6(\mathrm{mod}7)$. Consequentlygives an infinite set of positive integer solutions. Conversely these are the positive solutions in the indicated congruence classes, by the Chinese remainder theorem.