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Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.

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.

Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb{Z}\subseteq d\mathbb{Z}\cap h\mathbb{Z}$.

For the reverse inclusion, let $z=du=hv$ lie in this intersection of ideals. Since $d,h$ are coprime integers, Bézout's identity supplies $r,s\in \mathbb{Z}$ with $rd+sh=1$. Multiplying by $u$ givesThus $h\mid u$, and $dh\mid du=z$. ConsequentlyHere Bézout's identity follows from the allowed principal ideal domain property of $\mathbb{Z}$: the ideal $d\mathbb{Z}+h\mathbb{Z}$ is generated by their greatest common divisor, which is $1$.

Let $d=\gcd (c,n)$, and write $c=d{c}^{\prime }$, $n=d{n}^{\prime }$ with positive integers $c^{\prime },n^{\prime }$. If an integer $e>1$ divided both $c^{\prime }$ and $n^{\prime }$, then $de$ would divide both $c$ and $n$, contradicting the maximality of their greatest common divisor $d$. Thus $\gcd (c^{\prime },n^{\prime })=1$. Equivalently, Bézout's identity for $c,n$ can be divided by $d$ to express $1$ as an integer combination of $c^{\prime },n^{\prime }$.

The Euclidean algorithm repeatedly replaces $(r_{i-1},r_i)$ by $(r_i,r_{i-1}-q_i{r}_i)$, where the new remainder lies between $0$ and $r_i-1$. Common divisors are unchanged by this subtraction, and decreasing positive remainders ensure termination. The last nonzero remainder is the greatest common divisor.

The extended Euclidean algorithm also tracks coefficients $r_i=u_{i}a+v_{i}b$, beginning with coefficient pairs $(1,0)$ and $(0,1)$ and applying the same subtraction to the pairs. This invariant produces Bézout coefficients for the last nonzero remainder. Here the divisions areBack substitution givesThus $\gcd (580,252)=4$, with $m=10$ and $n=-23$ in Bézout's identity.

If $p\mid a$, there is nothing to prove. Otherwise the only possible positive common divisors of the prime $p$ and $a$ show that $\gcd (p,a)=1$. By Bézout's identity, $up+va=1$ for some integers $u,v$. Multiplying by $b$ gives $b=upb+vab$. Both terms on the right are divisible by $p$, the second because $p\mid ab$. Hence $p\mid b$. This proves Euclid's lemma, including when either factor is zero or negative.

For a formal group law over any commutative ring $R$, the invariant differential of a formal group law satisfies $[p{]}^{\ast }\omega =p\omega$. Its coefficient is a unit in a ring, so $[p{]}^{\prime }(T)\in pR[[T]]$. If $[p](T)=\sum a_m{T}^m$, this means $m{a}_m\in pR$. When $p\nmid m$, the integer Bézout's identity implies $a_m\in pR$, even if $p$ is a zero divisor. Put those coefficients into $pf(T)$ and the exponents divisible by $p$ into $g(T^p)$.

For positive integers $m,n,k$, the greatest common divisor scales by the common multiplier $k$. If $d=\gcd (m,n)$, then $kd$ is a common divisor of $km,kn$. Meanwhile Bézout's identity expresses $kd$ as an integer linear combination of $km,kn$, so their greatest common divisor divides $kd$. These two comparisons prove the identity without prime factorization.

For nonzero coprime integers $p,q$, the torus knot $T_{p,q}$ lies on a standard unknotted torus in the three-sphere and winds $p$ times meridionally and $q$ times longitudinally. Here a meridian of a solid torus goes around the small cross-section of the doughnut, while the longitudinal direction goes around its central hole. The integers count signed turns, so changing one sign gives the mirror of a link of the torus knot. In this article $p$ is always the meridional count and $q$ the longitudinal count.

An explicit parametric curve in three-dimensional Euclidean space iswhere $R>r>0$. Its cylindrical radius $\rho =\sqrt{x^2+y^2}$ satisfies $(\rho -R{)}^2+z^2=r^2$, the equation of the supporting torus. Its meridional angle is $pt$ and its longitudinal angle is $qt$: this checks the convention directly rather than relying on the picture. Its squared speed isIf two parameter values give the same point, their difference $\delta$ satisfies $p\delta ,q\delta \in 2\pi \mathbb{Z}$. Bézout's identity for the coprime integers $p,q$ then gives $\delta \in 2\pi \mathbb{Z}$. Thus the loop closes and is an injective function on the parameter circle, with nonzero derivative; it is a smooth embedding.

For example, $R=2$, $r=0.7$, $p=2$, $q=3$ produces a trefoil knot. During one circuit the meridional angle changes by $4\pi$ and the longitudinal angle by $6\pi$. The supporting torus is a guide for construction, not a constraint on later ambient isotopies: the knot may move off it. This family makes arithmetic visible as geometry: the coprime integers determine whether a single traversal visits the whole loop without repetition.

The coprime integers hypothesis matters. If the greatest common divisor is $d=\gcd (|p|,|q|)>1$, the same parametric curve retraces a single component $d$ times; it does not automatically draw $d$ different components. The corresponding torus link is instead the full set of points on the torus whose angles satisfy $q\varphi -p\theta \in 2\pi \mathbb{Z}$, with meridional angle $\varphi$ and longitudinal angle $\theta$. It has $d$ disjoint components, each with winding counts $(p/d,q/d)$. For instance, the torus link $T_{2,2}$ consists of two parallel $(1,1)$ loops and is a Hopf link, whereas substituting $(2,2)$ into ${\gamma }_{p,q}$ simply traces one of those loops twice.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]

Every nonempty subset of the natural numbers has a least element in their usual order. Applying this principle to positive integer linear combinations proves Bézout's identity. This principle for the usual natural-number order is distinct from the general well-ordering theorem asserting that every set can be equipped with some well-order.