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Write the cyclic factors additively. The order of a group element $(r,s)$ in a direct product of groups is the least common multiple of its component orders, since a power is the identity exactly when it is the identity in both coordinates.

In $C_2\times C_3$, $(1,1)$ has order $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt is well defined and multiplicative in additive notation; its kernel is zero because an integer divisible by both two and three is divisible by six. Equal finite orders then give bijectivity.

In $C_2\times C_4$, every element has order dividing four. A generator of $C_8$ has order eight, and isomorphisms preserve element orders, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, attained by a pair of generators. The product is cyclic of order $nm$ exactly when that number is $nm$. Thus cyclicity of a product of two finite cyclic groups givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

Use the agreed text encoding to split the plaintext into integer blocks $m_j\in \{0,\dots ,1072\}$. RSA encrypts each block by modular exponentiation:Because $1073=29\cdot 37$ is a product of distinct primes, its Euler totient function is $\phi (1073)=28\cdot 36=1008$. The inverse exponent follows fromThus a decryption exponent is 275, and the three recovered blocks areTo check every block, including blocks not coprime to the modulus, reduce modulo each prime. If $m_j$ is nonzero modulo 29, Fermat's little theorem gives $m_j^{3025}=m_j$ because $28\mid 3024$; if it is zero the equality is immediate. The same argument works modulo 37 because $36\mid 3024$. The Chinese remainder theorem then gives the equality modulo 1073. Finally reverse the agreed text encoding. The symbols $r_1,r_2,r_3$ are unspecified, so there are no numerical message blocks to evaluate further.

Euler's totient function isEuler's theorem states that $\gcd (a,n)=1$ implies $\boxed{a^{\phi (n)}\equiv 1(\mathrm{mod}n)}$. For $n>1$, let $r_1,\dots ,r_{\phi (n)}$ be the invertible residue classes modulo $n$. Multiplication by $a$ permutes these classes, since it has an inverse modulo $n$. Multiplying the resulting congruences givesEvery factor is a unit, so the product can be cancelled, proving the theorem. The case $n=1$ is immediate. For distinct prime numbers $p,q$, removing multiples of either prime from $1,\dots ,pq$ gives

For the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, and choose $e$ coprime to $\phi (N)$. Compute $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. The public key is $(N,e)$; the private exponent $d$ is retained secretly. Represent a message by $0\le m<N$, encrypt by $c\equiv m^e(\mathrm{mod}N)$, and decrypt by $m^{\prime }\equiv c^d(\mathrm{mod}N)$.

To prove correct decryption even for messages not coprime to $N$, write $ed=1+k(p-1)(q-1)$. Modulo $p$, if $p\mid m$ then $m^{ed}\equiv m\equiv 0$; otherwise Fermat's little theorem gives $m^{ed}\equiv m$. The same reasoning holds modulo $q$. The Chinese remainder theorem therefore gives $m^{ed}\equiv m(\mathrm{mod}N)$, and the chosen representative recovers $m$. Thus decryption works for every message residue, not merely the units.

Euler's totient function isEuler's theorem states that $\gcd (a,n)=1$ implies $\boxed{a^{\phi (n)}\equiv 1(\mathrm{mod}n)}$. For $n>1$, let $r_1,\dots ,r_{\phi (n)}$ be the invertible residue classes modulo $n$. Multiplication by $a$ permutes these classes, since it has an inverse modulo $n$. Multiplying the resulting congruences givesEvery factor is a unit, so the product can be cancelled, proving the theorem. The case $n=1$ is immediate. For distinct prime numbers $p,q$, removing multiples of either prime from $1,\dots ,pq$ gives

For the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, and choose $e$ coprime to $\phi (N)$. Compute $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. The public key is $(N,e)$; the private exponent $d$ is retained secretly. Represent a message by $0\le m<N$, encrypt by $c\equiv m^e(\mathrm{mod}N)$, and decrypt by $m^{\prime }\equiv c^d(\mathrm{mod}N)$.

To prove correct decryption even for messages not coprime to $N$, write $ed=1+k(p-1)(q-1)$. Modulo $p$, if $p\mid m$ then $m^{ed}\equiv m\equiv 0$; otherwise Fermat's little theorem gives $m^{ed}\equiv m$. The same reasoning holds modulo $q$. The Chinese remainder theorem therefore gives $m^{ed}\equiv m(\mathrm{mod}N)$, and the chosen representative recovers $m$. Thus decryption works for every message residue, not merely the units.

A primitive root modulo $p$ is an integer whose residue generates the multiplicative group ${\mathbb{F}}_p^{\times }$, equivalently has order $p-1$.

The Chinese remainder theorem states that, for pairwise coprime positive integers $m_1,\dots ,m_r$, the system $x\equiv a_i(\mathrm{mod}{m}_i)$ has a unique solution modulo $M={\prod }_i{m}_i$. For a constructive proof put $M_i=M/m_i$. Bézout's identity gives an integer $u_i$ with $M_i{u}_i\equiv 1(\mathrm{mod}{m}_i)$. Then $x={\sum }_i{a}_i{M}_i{u}_i$ has all the required residues. If two solutions exist, their difference is divisible by every $m_i$; pairwise coprimality makes it divisible by their product, proving uniqueness modulo $M$.

Here $3,5,7$ divide $p-1$, and none is zero modulo $p$. Choose a primitive root $g$ and exponents $a,b,c$ withUse the theorem with moduli $3,5,7$ to choose $r$ such that $r\equiv a(\mathrm{mod}3)$, $r\equiv b(\mathrm{mod}5)$, $r\equiv c(\mathrm{mod}7)$. Define nonzero residuesThe exponents are integers; negative ones are interpreted using inverses in ${\mathbb{F}}_p^{\times }$. Taking integer representatives now givesThe theorem aligns the exponent modular congruences, which is why a primitive root converts the three power conditions into a linear modular congruence problem.

Choose a nonzero quadratic nonresidue $c(\mathrm{mod}q)$. There is one because the squaring map on ${\mathbb{F}}_q^{\times }$ is two-to-one, so exactly half its elements are squares. The moduli $4,q,b$ are pairwise coprime, so the Chinese remainder theorem gives a positive integer $k$ satisfyingWhen $b=1$, the last condition is vacuous. In particular $k$ is odd and coprime to $qb$. Since $k\equiv 1(\mathrm{mod}4)$, Jacobi reciprocity givesAs $t$ is odd, multiplicativity of the Jacobi symbol now yields

Euler's theorem states that if $\gcd (a,m)=1$, then $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$, where $\varphi$ is Euler's totient function. This includes Fermat's little theorem when $m$ is prime.

For each prime-power factor $p_i^{e_i}$, the integer $\varphi (p_i^{e_i})$ divides the defined least common multiple $\lambda (m)$. Euler's theorem modulo that prime power therefore givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$. Equivalently the Chinese remainder theorem givesFor the specified composite integer,Raising $a^{80}\equiv 1(\mathrm{mod}561)$ to the seventh power provesThus 561 is a Carmichael number. We use exactly the source's definition of $\lambda$ as the least common multiple of prime-power totients; its validity here does not require replacing it by a smaller exponent at powers of two.

Choose a nonzero quadratic nonresidue $c$ modulo $q$. By the Chinese remainder theorem, choose a positive integer $k$ satisfyingThe moduli are pairwise coprime, and $k$ is coprime to $qb$. Jacobi reciprocity, with $k\equiv 1(\mathrm{mod}4)$, givesSince $t$ is odd, multiplying these symbols yieldsWhen $b=1$ its congruence imposes no restriction and its Jacobi factor equals one.

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.

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.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_1,\dots ,m_r$ and arbitrary integers $a_1,\dots ,a_r$, the simultaneous modular congruenceshave a solution, unique modulo $M={\prod }_i{m}_i$.

Choose $2017$ distinct prime numbers $q_1,\dots ,q_{2017}$. Their prime powers $q_i^{2017}$ are pairwise coprime. Apply the Chinese remainder theorem toFor the resulting integer $n$, each $n+i$ is divisible by $q_i^{2017}$, the required power of an integer greater than one. Adding a sufficiently large multiple of ${\prod }_i{q}_i^{2017}$ makes $n$ positive without changing any modular congruence. Thus such a positive $n$ exists, indeed infinitely many do. This is the general construction of consecutive integers divisible by prescribed prime powers.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

Write $|G|=pq$ with $p<q$. The Sylow theorems give $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. Since $p<q$, necessarily $n_q=1$. Its unique Sylow subgroup $N$ is a normal subgroup and has prime order $q$, soA Sylow subgroup $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. ThusThe semidirect product action is a group homomorphism $C_p\to \mathrm{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the Chinese remainder theorem; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a metacyclic group and hence a solvable group.

The primes $2,5$ both work, with the zero residue as a solution. For every other prime, multiplicativity of the Legendre symbol and quadratic reciprocity giveThe first factor is positive for $p\equiv 1,7(\mathrm{mod}8)$; the second is positive for $p\equiv 1,4(\mathrm{mod}5)$. They must have the same sign. The Chinese remainder theorem therefore yields

For a primitive nonprincipal Dirichlet character modulo $q$, finite Fourier inversion theorem and the primitive Gauss magnitude give this bound uniformly in the interval. Bound the exponential sums by $C\Vert a/q{\Vert }^{-1}$ and sum the resulting harmonic series. The general Gauss identity follows from the prime-power identity by the Chinese remainder theorem.

For an integer polynomial $F$ and squarefree $d$, let ${\rho }_F(d)$ count the roots of $F$ modulo $d$. The Chinese remainder theorem makes ${\rho }_F$ multiplicative, andThus $g(d)={\rho }_F(d)/d$ and $r_d=O({\rho }_F(d))$ define a sieve distribution.

Let $N=pq$ with primes $p\equiv 1(\mathrm{mod}12)$ and $q\equiv 5(\mathrm{mod}12)$. quadratic reciprocity makes three a square modulo $p$; choose $r^2=3$ and impose $a=-3+2r(\mathrm{mod}p)$, $a=1(\mathrm{mod}q)$ by the Chinese remainder theorem. The curve has good reduction at both primes. Indeed its discriminant is $-16a^2(4a+27)$; the first condition could make a factor vanish only at $p=3$ or 11, while the second gives $a(4a+27)=31$ and the given residue class excludes $q=31$.

Doubling $P$ gives $x(2P)-1=(a^2+6a-3)/4$. This vanishes modulo $p$ and equals one modulo $q$. Modulo $p$ one also has $y(2P)=-1$, so $2P=-P$ and $3P$ is the identity. Modulo $q$ the addition is finite. Thus computing $3P$ modulo $N$ encounters a denominator whose greatest common divisor with $N$ is exactly $p$, revealing the factor. The numerator difference is minus two modulo $p$, so the failed inverse cannot be cancelled. This is a construction of successful elliptic curve method data from prescribed local group behavior.