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For a generator of a group $g$ of order $M$ and $x=g^y$, the function $f(a,b)=g^a{x}^{-b}$ hides the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$ of ${\mathbb{Z}}_M^2$. Each fiber is an affine coset $\{(yb+k,b):b\in {\mathbb{Z}}_M\}$. Measuring the function register gives its normalized coset state. Applying the positive-exponent quantum Fourier transform to both input registers gives amplitudeThe finite geometric series vanishes off $c_2=-y{c}_1$ and equals $M$ on that line. Thus each allowed output pair has probability $1/M$, independently of the coset offset. If $\gcd (c_1,M)=1$, a modular inverse recovers $y=-c_2{c}_1^{-1}$.

Taking the real and imaginary parts of the finite geometric series ${\sum }_{m=1}^N{e}^{im\theta }$ yieldsThese formulas apply when $\theta \notin 2\pi \mathbb{Z}$. At multiples of $2\pi$, the sums are directly $N$ and $0$. Complex exponentials turn two real trigonometric summations into one finite geometric series.

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 Roth theorem on three-term arithmetic progressions says that, for each fixed $\delta >0$, a sufficiently long interval cannot have a subset of density of a finite subset at least $\delta$ with no nonconstant three-term arithmetic progression. Equivalently, the maximum size of a three-term-progression-free subset of $[N]=\{1,\dots ,N\}$ is $o(N)$. Here is a density increment proof, including its analytic ingredients.

Write $e(t)=\exp (2\pi it)$ and embed $I=[N]$ into the cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$ of order $M$. For normalized Fourier analysis on a finite abelian group take $\hat{h}(r)=M^{-1}{\sum }_{x}h(x)e(-rx/M)$. The orthogonality of complex exponentials follows from a finite geometric series: its average is zero for a nonzero frequency and one for frequency zero. Expanding in these linear phases consequently proves both Fourier inversion on a finite group and the Parseval identity on a finite group. In particular,The second identity follows by summing the frequency expansion over $x,d$: the two constraints leave precisely the frequencies $(r,-2r,r)$. Since $M$ is odd, multiplication by two permutes the frequencies. Also a congruence $x+z=2y$ with $x,y,z\in I$ is an equality in the integers, since $|x+z-2y|<M$.

Let $A\subseteq I$ have subset density $\alpha =|A|/N\ge \delta$, and suppose it contains no nonconstant three-term arithmetic progression. Its balanced indicator function of a finite subset is $f=1_A-\alpha 1_I$, extended by zero outside $I$. Then $\hat{f}(0)=0$. Counting endpoints with the same parity givesIf $N\ge 4{\alpha }^{-2}$, their difference in absolute value is at least ${\alpha }^3{N}^2/(4M^2)$. Telescope the difference of the products into three terms, each containing $f$ and two factors chosen from $1_A,\alpha 1_I$. Put $s={\max }_r|\hat{f}(r)|$. The Cauchy-Schwarz inequality and the Parseval identity on a finite group bound each term by $s\alpha N/M$, since the squared normalized $L^2$ norm of either other factor is at most $\alpha N/M$. Therefore some nonzero frequency $r$ satisfiesThis is the Fourier detection of a progression-free subset of an interval step; it has just been proved rather than assumed.

We now turn this Fourier coefficient on a finite abelian group into a density increment. Set $Q=\lfloor \sqrt{N}\rfloor$. Partition the unit interval into $Q$ equal pieces and place the $Q+1$ fractional parts of $0,r/M,\dots ,Qr/M$ in them. The pigeonhole principle gives $1\le q\le Q$ with $\Vert qr/M\Vert \le 1/Q$. This proves the particular Dirichlet approximation theorem needed here. Put $\kappa ={\alpha }^2/(1000\pi )$ and $L=\lfloor \kappa Q\rfloor$. For sufficiently large $N$ depending only on $\delta$, $L\ge \kappa Q/2$ and $2L\le Q$. Each residue chain modulo $q$ in $I$ has at least $Q$ elements. Split each such chain into arithmetic progressions with between $L$ and $2L-1$ elements, absorbing its final short remainder into the previous block.

The linear phase changes by at most $4\pi L/Q\le {\alpha }^2/250$ within any block. Freezing the linear phase at the first element of each block therefore changes the preceding correlation by at most ${\alpha }^{2}N/250$, because $|f|\le 1$. The triangle inequality givesThe block sums of the balanced indicator function of a finite subset add to zero, so their positive total is half their absolute total. Some block consequently satisfiesHere $c>0$ is absolute, and shrinking it absorbs the floors. An affine parametrization of $P$ identifies $A\cap P$ with a subset of $[|P|]$ and preserves arithmetic progressions.

Repeat this Roth density-increment step. Every step raises the subset density by at least ${\delta }^2/144$, so fewer than $T=\lceil 144/{\delta }^2\rceil +1$ steps are possible. All the lengths satisfy $N_{j+1}\ge c{\delta }^2\sqrt{N_j}$ while above a threshold depending only on $\delta$. Choose the initial $N$ so large that this lower recursion stays above that threshold for $T$ steps; this is possible by working backwards through finitely many squarings. We would then force a subset density greater than one. Thus every fixed positive density eventually forces a nonconstant three-term arithmetic progression. For an infinite subset of the positive integers with positive upper asymptotic density, apply the finite conclusion along initial intervals whose subset density is bounded away from zero; this gives the usual infinite formulation as well.