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Use the Euclidean algorithm: $23=18+5$, $18=3\cdot 5+3$, $5=3+2$, and $3=2+1$. Back-substitution gives the Bezout identity $1=9\cdot 18-7\cdot 23$. Multiplying by $101$ gives one solution of the linear Diophantine equation, and shifting the coefficients by multiples of $(23,-18)$ reduces it toIndeed, $18\cdot 12+23(-5)=216-115=101$. More generally all solutions are $x=12+23k$, $y=-5-18k$, $k\in \mathbb{Z}$: subtracting this particular solution gives $18(x-12)=-23(y+5)$, and the coprime integers $18,23$ force $23\mid (x-12)$.

The Euclidean algorithm givesThus the greatest common divisor is $d=7$. The extended Euclidean algorithm reverses these divisions:This supplies the Bezout identity with $x=8$, $y=-11$.

To obtain all solutions of a linear Diophantine equation, subtract this particular solution from any other and divide by $7$. The result is $29(x-8)+21(y+11)=0$. Since $29$ and $21$ are coprime integers, $21$ divides $x-8$, givingEvery displayed pair satisfies the original linear Diophantine equation, so the parametrization is exhaustive.

For the modular arithmetic calculation, $21\cdot 18=378\equiv 1(\mathrm{mod}29)$, so $18$ is the modular inverse of $21$. Multiplication gives $n\equiv 25\cdot 18\equiv 15(\mathrm{mod}29)$. The admissible integers are $15+29t$ for $0\le t\le 68$, ending at $1987$; the next is $2016$. There are $\boxed{69}$ such integers.

Let $D$ be a finite integral domain. For any $a\ne 0$, multiplication $m_a:D\to D$ is injective: $ax=ay$ implies $a(x-y)=0$, hence $x=y$ because there are no zero divisors. An injective map of a finite set to itself is surjective, so $ab=1$ for some $b$. Every nonzero element is invertible; therefore a finite integral domain is a field.

For an irreducible polynomial $f\in F[X]$, consider a nonzero class $[g]$ in the quotient ring $F[X]/(f)$. Since $f$ does not divide $g$, irreducibility gives $\gcd (f,g)=1$. The Euclidean algorithm in the polynomial ring gives polynomials $u,v$ with $uf+vg=1$. Modulo $(f)$, $[v][g]=1$. Thus the quotient by an irreducible polynomial is a field.

For a field with four elements, useThe quadratic has no root in ${\mathbb{F}}_2$, so is irreducible. Write $\alpha =[X]$. Polynomial division shows the elements are $0,1,\alpha ,\alpha +1$, with ${\alpha }^2=\alpha +1$. Its multiplication table is

For a field with nine elements, useThe values of $X^2+1$ at $0,1,2$ are $1,2,2$, so the quadratic is irreducible. Every class is uniquely $a+b\beta$, $a,b\in {\mathbb{F}}_3$, with ${\beta }^2=-1=2$. These nine classes form a finite field; multiplication is $(a+b\beta )(c+d\beta )=(ac-bd)+(ad+bc)\beta$, with coefficients modulo three.

Use the positive-sign quantum Fourier transformOrthogonality of the finite-group characters makes $F_N$ a unitary operator. Here $r$ is the least positive period. The injectivity within one period implies $r\mid N$: write $N=qr+s$, $0\le s<r$. Periodicity gives $f(s)=f(N)=f(0)$, and injectivity forces $s=0$. Put $L=N/r$.

Start with two available computational basis states $|0⟩|0⟩$. Apply $F_N$ to the first register and query the modular-addition quantum oracle:Measure the second register. For some $t\in \{0,\dots ,r-1\}$, the first register becomes the normalized coset stateThe quantum Fourier transform of a periodic coset state isIndeed, the inner geometric sum vanishes unless the Fourier label is a multiple of $N/r$. Measuring that label gives $j=sN/r$ for a uniformly random $s$. Compute, by the Euclidean algorithm,Thus exact period recovery from a Fourier sample succeeds whenever $s$ is coprime to $r$, andHere $\phi$ is the Euler totient function. The classical number-theory input is the totient lower bound from the Mertens product: for sufficiently large $m$, $\phi (m)/m\ge c/\log \log m$ for an absolute $c>0$, with the finitely many small cases handled separately. For $r=1$ recovery is certain; for $r=2$ its probability is $1/2$. The useful asymptotic probability statement is a lower bound, hence $\Omega$ notation; it can be much larger for particular periods, for example a prime period.

There is one oracle query, two Fourier transforms and two measurements. The final greatest common divisor and division use polynomial time in $\log N$. This meets the specified primitive-operation model without assuming that an arbitrary-$N$ Fourier gate is free in another gate model. If a constant success probability is wanted, repeat independently $O(\log \log N)$ times and return the least common multiple of the candidates: each candidate divides $r$, and one successful sample makes that least common multiple exactly $r$.

Write $Q=2^n$, $t=(p-1)(q-1)$ and $a=t/Q$. The easy starting state is uniform over all $n$-bit labels, so its good fraction is $a$, not $t/N$. For distinct primes one has $a\ge 1/4$. If both primes are odd, $t/N\ge (1-1/3)(1-1/5)=8/15$ and $N/Q\ge 1/2$, giving $a\ge 4/15$. If one prime is two, the other is an odd prime $q$; since $2q$ has $n$ bits, $q>2^{n-2}$, and $t=q-1\ge 2^{n-2}=Q/4$.

The uniform coprime state for a semiprime can now be prepared by one exact Grover rotation. SetPrepareUse the good subspace spanned by $|k⟩|1⟩$ with $1\le k<N$ and $\gcd (k,N)=1$. Its squared overlap with $|\Psi ⟩$ is exactly $a\lambda =1/4$. The Euclidean algorithm supplies a reversible computation of the membership predicate, including the range check; condition a sign flip on membership and the extra qubit being one, then uncompute the workspace. This implements $S_G=I-2{\Pi }_G$.

The starting-state reflection $D_{\Psi }=2|\Psi ⟩⟨\Psi |-I$ uses the inverse of its known preparation and a reflection on the all-zero state. Applying $D_{\Psi }{S}_G$ once invokes exact amplitude amplification at $\theta =\pi /6$, and givesThe extra qubit factors off, and the arithmetic workspace is returned to zero. Although the stated range includes $N$, that label is not coprime to itself, so the predicate $k<N$ is equivalent and avoids admitting labels outside the intended range.

The supplied $p,q$ determine $t$ in polynomial time; no factoring procedure is needed. Binary arithmetic, range comparison, the reversible greatest common divisor, the two starting-state reflections and the controlled sign flip all have polynomial-size circuits. Thus the construction takes polynomial time in $n$ in the ideal model allowing the specified one-qubit state preparation. Exactness uses the rotation with known amplitudes $\sqrt{\lambda },\sqrt{1-\lambda }$; with a fixed finite approximate gate library one obtains arbitrary accuracy with precision overhead, rather than an automatic promise of exact state preparation. In the ideal model the preparation succeeds with certainty, without rejection sampling. If desired, the extra $|1⟩$ can be reset by a Pauli X gate.

Let $L=\lceil {\log }_{2}N\rceil$. In the Shor algorithm, first use classical arithmetic to deal with even $N$ and perfect powers, and to reject prime inputs. Perfect-power recognition and primality testing have polynomial-time classical algorithms. Factoring a perfect power $b^k$, with $b,k>1$, already gives a nontrivial factor $b$. We may therefore consider an odd composite $N$ with at least two distinct prime factors.

Choose $\alpha$ uniformly from $1<\alpha <N$ and compute its greatest common divisor with $N$ using the Euclidean algorithm. A proper nontrivial divisor ends the computation. Otherwise $\alpha$ is a unit, and the remaining task is quantum order finding. SetPrepare a uniform quantum superposition in a $t$-qubit register and use quantum modular exponentiation in a second register to obtainRepeated squaring and reversible modular multiplication implement this step with polynomially many quantum gates in $L$. Apply the inverse quantum Fourier transform to the first register and perform a quantum measurement in the computational basis, obtaining $j$. Measuring the second register first, and Fourier-transforming the resulting periodic coset, gives the same distribution of $j$.

Here is a quantitative quantum Fourier sampling bound for order finding. If $r={\mathrm{ord}}_N(\alpha )$, multiplication by $\alpha$ cycles its $r$ orbit states. The Fourier eigenstates of that cycle have eigenvalues $e^{2\pi is/r}$, and the initial state $|1⟩$ has squared overlap $1/r$ with each, for $0\le s<r$. ThereforeConditioned on a particular eigenphase $s/r$, its nearest integer outcome $j_s$ obeys $|j_s/Q-s/r|\le 1/(2Q)$ and has probability at least $4/{\pi }^2$. Indeed the absolute amplitude is $|\sin (\pi Q\delta )/(Q\sin (\pi \delta ))|$, where $\delta =s/r-j_s/Q$. For $|Q\delta |\le 1/2$, the numerator is at least $2Q|\delta |$ and the denominator at most $\pi Q|\delta |$; the limiting value at $\delta =0$ is one.

If $\gcd (s,r)=1$, the fraction $s/r$ is reduced, and $r<N$ givesUse continued-fraction recovery in quantum order finding. The relevant approximation theorem says that reduced fractions satisfying $|a/b-p/q|<1/(2q^2)$ are continued fraction convergents of $a/b$. For an input rational of $O(L)$ digits, their $O(L)$ candidates can be enumerated classically in $O(L^3)$ time, and their denominators are at most the reduced input denominator. This is the supplied continued-fraction result; discard candidate denominators exceeding $N$. A zero measurement is treated as a failed order-finding sample.

For each remaining even candidate denominator $q$, compute $z={\alpha }^{q/2}\mathrm{mod}N$ by repeated squaring, and try $\gcd (z-1,N)$ and $\gcd (z+1,N)$. Return only a divisor $d$ with $1<d<N$; otherwise repeat the quantum sample or choose another $\alpha$. This candidate-denominator gcd post-processing cannot return an incorrect factor, since its output is checked directly. When the recovered denominator is the true even multiplicative order and the square root of one is nontrivial, part (i) guarantees success. Checking ${\alpha }^q\equiv 1(\mathrm{mod}N)$ before trying the gcds is also a valid, more restrictive convention; the single-run distinction matters in part (iii).

Two further standard quantitative facts explain why repetitions remain efficient. First, for a uniformly chosen unit modulo an odd integer with at least two distinct prime divisors, the probability of a useful unit in Shor factorization is at least $1/2$: its true multiplicative order is even and ${\alpha }^{r/2}\not\equiv -1(\mathrm{mod}N)$. To see the source of this bound, the Chinese remainder theorem for unit groups makes the prime-power components independent. Each odd-prime-power unit group is cyclic. If its 2-primary order is $2^v$, the probability of the largest possible value of $v$ is $1/2$, and the probability of any specified smaller value is at most $1/2$. Failure requires all components to have the same $v$: either all zero, making $r$ odd, or all the same positive value, making the halfway power $-1$ in every component. Conditioning on one component, the chance another component matches it is at most $1/2$, so total failure is at most $1/2$.

Second, the fraction of phase numerators coprime to $r$ is $\phi (r)/r$, where $\phi$ is the Euler totient function. The elementary totient-ratio lower bound suffices: if the distinct prime divisors are $p_1<\cdots <p_k$, then $p_j\ge j+1$ and $k\le {\log }_{2}r$. The totient product formula therefore givesThus for a good $\alpha$, the probability of sampling a coprime phase numerator with the required accuracy is at least $(4/{\pi }^2)\phi (r)/r$. This is inverse-polynomial in the input length, so polynomially many repetitions achieve any fixed success probability. The quantum Fourier transform uses $O(t^2)$ elementary rotations and Hadamard gates in the ideal gate description, and the other quantum and classical stages are polynomial in $L$.

The reduction is therefore polynomial in the bit length:A small denominator need not itself be the true multiplicative order; verification of the final factor is essential, and no guarantee is claimed for every individual sample.

Use the positive-exponent quantum Fourier transform convention $F_N|x⟩=N^{-1/2}{\sum }_t{e}^{2\pi ixt/N}|t⟩$. The period of a function here is the least positive translation leaving the function unchanged on the cyclic group ${\mathbb{Z}}_N$. The injective promise means $f(x)=f(x^{\prime })$ exactly when $x\equiv x^{\prime }(\mathrm{mod}r)$. In particular $r\mid N$: the subgroup of period translations is generated by $r$, and wrapping the argument by $N$ preserves its value. This is exact quantum period finding on a finite cyclic group, rather than the truncated, nondividing-period version requiring continued fractions.

Prepare a uniform superposition state in the first quantum register using $F_N|0⟩$, and perform a reversible computation of the function into a second quantum register, cleaning its workspace by uncomputation:A quantum measurement in the computational basis of the second quantum register leaves a periodic coset state in the first:For the quantum Fourier transform of this periodic coset state, the finite geometric series vanishes unless the measured label is $t=mN/r$. On those labels its normalized amplitude is $r^{-1/2}{e}^{2\pi iam/r}$. Thus $m$ has the uniform distribution on a finite set on $\{0,\dots ,r-1\}$, irrespective of $a$.

Compute the greatest common divisor by the Euclidean algorithm, and set $q=N/\gcd (t,N)$, with $\gcd (0,N)=N$. Since $t=mN/r$,The exact period recovery from a Fourier sample is therefore certified by testing $f(q\mathrm{mod}N)=f(0)$. Every candidate $q$ divides $r$ and lies between $1$ and $r$, so the injective promise makes that equality equivalent to $q=r$. This includes $r=N$, where the test uses input $0$, and $r=1$, which always succeeds. The test does not assert that an arbitrary equality of function values certifies an arbitrary period; the Fourier-derived divisor restriction is essential.

The concise result isHere $\phi$ is the Euler totient function, with $\phi (1)=1$. The totient lower bound from the Mertens product gives $\phi (r)/r=\Omega (1/\max \{1,\log \log N\})$ for $r\le N$. Each trial uses polynomial time quantum Fourier transforms, function evaluations and Euclidean algorithm arithmetic in $\log N$. Repeating $O(\max \{1,\log \log N\})$ independent trials gives constant success probability; the herald accepts only correct answers.

The printed asymptotic bound has the wrong direction for this argument. In Big O notation, the stated upper bound on the number of coprime integers supplies no success lower bound and is itself false uniformly: for prime $N$, $\phi (N)=N-1$, not $O(N/\log \log N)$. Likewise the single-trial success need not be $O(1/\log \log N)$, since it tends to one for $r=N$ prime. The meaningful corrected guarantee is the lower bound above. If an upper-bounded success rate is literally required, one may deliberately retain an accepted result with an independent probability proportional to $1/\max \{1,\log \log N\}$; this discards useful successes and is unnecessary for efficient recovery.