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Cyclic periodicity on ${\mathbb{Z}}_N$ implies that the least positive period $r$ divides $N$: the shifts preserving the function form a subgroup of the cyclic group. Injectivity within one period says that distinct residue classes modulo $r$ have distinct function values. Write $L=N/r$.

Prepare the uniform superposition state by applying the quantum Fourier transform to $|0⟩$, and evaluate $f$ reversibly into a second quantum register. This is efficient because the given classical Boolean circuit can be converted into a reversible circuit, with its workspace cleaned by uncomputation. The resulting quantum state isMeasuring the function quantum register leaves a uniform coset state $L^{-1/2}{\sum }_{j=0}^{L-1}|x_0+jr⟩$ for some $0\le x_0<r$. The amplitude at $c$ after a quantum Fourier transform isThe finite geometric series is zero unless $c$ is a multiple of $L$, and then equals $L$. Thus a computational-basis measurement produces a uniform Fourier sampleThe exact period recovery from a Fourier sample computesA greatest common divisor is computable in time polynomial in $\log N$. The candidate is always a divisor of the true period, and equals it precisely when $s$ is coprime to $r$.

For the success flag, compute $f(\hat{r}\mathrm{mod}N)$ and compare with $f(0)$. If $\hat{r}<r$, injectivity within the first period makes these unequal. If $\hat{r}=r$, periodicity makes them equal, including $r=N$. Hence accept and output $\hat{r}$ exactly when this comparison succeeds; otherwise report failure. This is heralded exact quantum period finding when the period divides the register size, with no false acceptance. The case $r=1$ succeeds even on the sample $c=0$.

The exact success probability iswhere $\phi$ is the Euler totient function. For $r\mid N$, the prime-product formula gives $\phi (r)/r\ge \phi (N)/N$. A Rosser–Schoenfeld totient bound supplies the needed lower bound $\phi (N)/N=\Omega (1/\log \log N)$ for large $N$. Every single attempt, including verification, has the assumed polynomial-in-$\log N$ gate cost. Repeating $O(\log \log N)$ times gives constant success probability with a certified result whenever one is found.

The printed $O$ in the totient hint should be a lower-bound statement, $\Omega$. Literally, $\phi (N)=O(N/\log \log N)$ is false: at primes $N$, $\phi (N)=N-1$. Likewise the exact single-sample success probability need not be $O(1/\log \log N)$; for prime period it tends to one. The useful intended guarantee is the displayed inverse-log-log lower bound, not a universal upper bound. If an algorithm satisfying the literal success upper bound is desired, one can additionally accept an otherwise certified run only when $k=\lceil \log \log N\rceil$ independent fair bits are all zero. This retains a detectable failure flag and polynomial runtime, and has success at most $2^{-k}=O(1/\log \log N)$, but deliberately weakens the useful guarantee. The standard unsuppressed algorithm above gives the stronger intended performance. Small $N$ can be handled directly. Primary evidence for the totient lower bound is Theorem 15, equations (3.41)–(3.42), in denisevellachemla.eu/Rosser-Schoenfeld-1962.pdf .

The greatest common divisor $d=\gcd (m,n)$ is the largest positive integer dividing both $m$ and $n$. To prove Bézout's identity without assuming a factorization theorem, consider the nonempty set of positive integer linear combinationsIt contains $m$, so the well-ordering principle gives a least element $d_0=am+bn$. By Euclidean division, write $m=q{d}_0+r$ with $0\le r<d_0$. Thenis another integer linear combination. If $r>0$, it contradicts minimality, so $r=0$ and $d_0\mid m$. Applying the same argument to $n$ gives $d_0\mid n$. Conversely, every common divisor divides every integer linear combination and therefore divides $d_0$. In particular every positive common divisor is at most $d_0$. Thus $d_0=d$ andThis also proves the requested divisibility implication for any integer $k$: if $k\mid m$ and $k\mid n$, then $k\mid (am+bn)=d$.

For the scaling identity for greatest common divisors, put $D=\gcd (km,kn)$ with $k>0$. Since $kd$ divides $km$ and $kn$, $D\ge kd$. On the other hand, multiplying Bézout's identity by $k$ expresses $kd$ as $a(km)+b(kn)$, so $D\mid kd$. Positivity gives $D\le kd$. Therefore without using prime factorization,

The PDF's small superscript is essential: $f(x)=a^x\mathrm{mod}N$. Interpret $r$ as the least positive period, the multiplicative order of $a$ modulo $N$. If one permitted an unspecified nonminimal period, the answer would not be identifiable; for example, $a=1,N=8$ gives the same constant function with period one or period eight. Under the promised periodic modular-exponential interpretation, $a^r\equiv 1(\mathrm{mod}N)$, so $a$ is a unit. ThusIn particular, different residue classes modulo $r$ have different function values, which is necessary for the usual quantum period finding coset argument.

Prepare $N^{-1/2}{\sum }_x|x⟩|1⟩$ by applying the quantum Fourier transform to the first register. Use repeated squaring to precompute $a^{2^j}\mathrm{mod}N$, then multiply the value register by that known number controlled on bit $j$ of $x$. These controlled modular multiplications are reversible because $a$ is a unit; their inverses use the modular inverses of the same constants. This produces $N^{-1/2}{\sum }_x|x⟩|a^x\mathrm{mod}N⟩$ with a polynomial number of the assumed arithmetic operations. Known work registers can be uncomputed.

Measure the value register. Since $r\mid N$, its fiber is exactly one coset, and the first register becomesThe quantum Fourier transform of a periodic coset state is supported uniformly on the $r$ outcomes $c=jN/r$, $0\le j<r$. Indeed the inner geometric series ${\sum }_{\ell =0}^{N/r-1}{e}^{2\pi i\ell rc/N}$ is zero unless $(N/r)\mid c$, and each allowed amplitude has modulus $1/\sqrt{r}$.

Take two independent Fourier samples $c_1,c_2$ and returnThis is two-sample exact period recovery. The Chinese remainder theorem makes divisibility of independent uniform residues independent across the different prime divisors of $r$. For each such prime $\ell$, both $j_1$ and $j_2$ are divisible by $\ell$ with probability ${\ell }^{-2}$. ThereforeThe number-theoretic facts used here are the Chinese remainder theorem, the Euler product for the Riemann zeta function at two, and $\zeta (2)={\pi }^2/6$. The candidate can also be certified by $a^{\hat{r}}\equiv 1(\mathrm{mod}N)$: it divides the least period, so this equality holds exactly when it is the full period. Each preparation, reversible evaluation, QFT, measurement and greatest common divisor calculation has the assumed or standard polynomial time cost in $\log N$. Two runs suffice for the desired constant success probability. Quantum period finding recovers the least period with probability at least $6/{\pi }^2$ in polynomial time.

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.

Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

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.

For an additive arithmetic function $f$, one standard form of the Turán-Kubilius inequality usesIt states, with an absolute constant,For real $f$, this also bounds the variance under the discrete uniform distribution, because centering at the actual expectation minimizes the mean square.

For the prime omega function, $f(p^k)=1$. The Mertens theorem for reciprocal primes states ${\sum }_{p\le N}1/p=\log \log N+O(1)$. The contribution of higher prime powers is bounded by ${\sum }_{p}1/(p(p-1))\le 1$, as proved in 1(a). It follows that $A_{\omega }(N)=\log \log N+O(1)$ and $B_{\omega }(N{)}^2\ll \log \log N$. Writing $L=\log \log N$, the Turán-Kubilius inequality and Chebyshev inequality therefore yieldThe bounded error in $A_{\omega }-L$ is absorbed by, for example, using the threshold $0.005L$ around $A_{\omega }$ for sufficiently large $N$.

For the multiplication table problem, call a pair $(a,b)\in [N{]}^2$ good when $\omega (a),\omega (b)\ge 0.99L$ and $\omega (\gcd (a,b))\le 0.1L$. The preceding Chebyshev inequality excludes only $O(N^2/L)$ pairs by either of the first two conditions. For the greatest common divisor condition, linearity of expectation under uniform sampling of the two coordinates givesMarkov inequality therefore excludes only another $O(N^2/L)$ pairs.

The prime factors of a product form the union of those of its factors. Consequently, for every good pair,On the other hand, apply the proved concentration estimate at $N^2$. Since $\log \log (N^2)=L+\log 2$, for large $N$ every integer $m\le N^2$ with $\omega (m)\ge 1.88L$ is exceptional there. There are only $O(N^2/L)$ such integers. Products of good pairs lie in this exceptional set; products having only bad representations are at most as numerous as the bad pairs. This proves the multiplication-table upper bound, using distinct prime factors alone:

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.

For coprime integers $\alpha ,N$, the multiplicative order is the smallest positive exponentIt exists because multiplication by $\alpha$ is an element of the finite group of units modulo an integer. For the example, the successive residues areNone of the first three is one, so the order is four.

The factor extraction from an even modular order uses the factorization ${\alpha }^r-1=({\alpha }^{r/2}-1)({\alpha }^{r/2}+1)$. If the multiplicative order $r$ is even, put $z={\alpha }^{r/2}\mathrm{mod}N$. Minimality of the multiplicative order excludes $z=1$; we also require $z\ne -1(\mathrm{mod}N)$. Thus $z^2\equiv 1(\mathrm{mod}N)$ is a nontrivial square root of one. Neither $z-1$ nor $z+1$ is divisible by all of $N$. If either were coprime to $N$, its inverse modulo $N$, applied to $(z-1)(z+1)\equiv 0$, would make the other divisible by $N$, a contradiction. Hence the greatest common divisorsare proper nontrivial factors. For odd $N$, $(z-1)$ and $(z+1)$ have no common prime divisor of $N$, so these two divisors are complementary. If the initial $\alpha$ is not coprime to $N$, a proper $\gcd (\alpha ,N)$ already supplies a factor without quantum order finding.

For $N=15$ and $\alpha =7$, $r=4$ and $z=7^2\mathrm{mod}15=4$, which is neither $1$ nor $14$. The factors areThe necessary conditions for this order-based reduction are a unit $\alpha \mathrm{mod}N$, even true multiplicative order, and ${\alpha }^{r/2}\not\equiv -1(\mathrm{mod}N)$.

Every reachable state $i$ reaches zero deterministically in $i$ steps. For $j\in E$, choose a supported lifetime $k>j$; zero can jump to $k-1$ and then count down to $j$. Thus the chain is an irreducible Markov chain on $E$. Its positive-probability first return lengths are exactly the support of $Y_1$; all other return lengths are sums of such lengths. Their greatest common divisor is therefore one by assumption, so the chain is an aperiodic Markov chain.

The equilibrium residual-life distribution isIt is a probability distribution because the tail-sum identity gives ${\sum }_{j\ge 0}\mathbb{P}(Y_1>j)=\mathbb{E}{Y}_1=\mu$. With zero masses outside $E$,so it is a stationary distribution. This also displays why finite mean is needed for normalization.

Use the countable-state Markov chain convergence theorem: an irreducible, aperiodic, positive recurrent discrete-time chain on a countable state space has a unique stationary distribution and $P^n(i,j)\to {\pi }_j$ for every pair of states. All hypotheses were verified, including positive recurrence in part (b). Since $X_0=0$ and $\mathbb{P}(Y_1>0)=1$,The finite-state convergence theorem alone would not suffice when the lifetimes are unbounded. The gcd hypothesis is also necessary for this ordinary limit: deterministic lifetime 2, for example, produces alternating renewal probabilities rather than convergence.

A simultaneous transit in a resonant chain requires both conjunction patterns to admit the observer's longitude at the same time. In the leading mean-conjunction geometry put $x=\Lambda -{\varpi }_c$. The necessary phase compatibility at symmetric centres isLet $d=\gcd (q,n)$. Eliminating $x$, and using that $q\ell -nk$ ranges over multiples of $d$, gives the exact modular compatibility criterionHere $\gcd$ denotes the greatest common divisor. For the orders $q,n\in \{1,2,3\}$ above, the two phase sets intersect only whenThe common directions are the two quadratures relative to periapsis. In reduced resonances this requires $p$ and $m$ odd; otherwise the apparent second-order ratios reduce to different resonance orders. There is no additional restriction on those odd values from angular compatibility alone. A shared direction and suitable temporal phase can be chosen, and the rational period ratios allow its recurrence.

This is a potential low-amplitude configuration, not a guarantee of an observed triple exoplanet transit. The observer must lie in the common orbital plane and near an allowed direction. Finite stellar radii, nonzero libration amplitudes of a resonant argument, apsidal motion and asymmetric centres broaden or change the possibilities. At finite $e_c$, true simultaneous alignment obeys the more precise equations $(p+q)y-px={\varphi }_{bc}$ and $(m+n)x-my={\varphi }_{cd}$, with $y=M_c(x)$. Thus the order-only conclusion is not a universal necessary condition for every eccentric resonant chain.

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.

A finite field extension $L/K$ is a separable field extension if every element of $L$ has a minimal polynomial over $K$ with distinct roots in an algebraic closure. The primitive element theorem says that a finite separable field extension $L/K$ is $K(\alpha )$ for some $\alpha \in L$.

If $K$ is finite, then so is $L$. The multiplicative group of a finite field is cyclic, so a generator of $L^{\times }$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a finite abelian group its exponent is attained as an element order, while the polynomial root bound forces its size to be at most that exponent.

If $K$ is infinite, first consider $L=K(a,b)$. The minimal polynomials of $a$ and $b$ have distinct roots $a_i,b_j$ in an algebraic closure. Choose $c\in K$, $c\ne 0$, avoiding the finitely many values for which $a_i+c{b}_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha =a+cb$. Over $K(\alpha )$, the polynomials $f(T)$ and $g((\alpha -T)/c)$ have precisely one common root, $a$. Their monic greatest common divisor is therefore $T-a$, so $a\in K(\alpha )$ and $b=(\alpha -a)/c\in K(\alpha )$. Thus $K(a,b)=K(\alpha )$. Induct over finitely many generators of the finite separable field extension. This proves every finite separable extension is simple.

The Euclidean algorithm repeatedly replaces a pair $(a,b)$ by $(b,a-qb)$ without changing its greatest common divisor. Reversing the divisions expresses $\gcd (a,b)$ as an integer linear combination $\alpha a+\beta b$, which is Bézout's identity.

Here $3=3(15)-2(21)$, and thereforeOne solution is $(x,y,z)=(36,-24,-1)$. It is not unique: for every $t\in \mathbb{Z}$, $(x,y,z)=(36+7t,-24-5t,-1)$ is another solution.

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

A nonzero element of a free abelian group is primitive when it is part of an integral basis, equivalently when some homomorphism to $\mathbb{Z}$ takes it to one. Its coordinates have greatest common divisor one. This lattice meaning differs from a primitive homology class in the coalgebra sense.

A Pythagorean triple consists of three positive integers satisfying $x^2+y^2=z^2$. It is primitive when their greatest common divisor is one. Reducing their square numbers modulo $3$ shows that at least one leg is divisible by $3$; reducing modulo $4$ shows that both legs cannot be odd.

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.