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The Fermat-Euler theorem states that for an integer $a$ and a positive integer $n$ with $\gcd (a,n)=1$,where the Euler totient function $\phi (n)$ counts the residue classes coprime to $n$. For $n\ge 2$, choose a reduced residue system $r_1,\dots ,r_{\phi (n)}$. Multiplication by the unit modulo n $a$ preserves coprimality and is injective on these residue classes: $a{r}_i\equiv a{r}_j(\mathrm{mod}n)$ implies $r_i\equiv r_j(\mathrm{mod}n)$, because $a$ has a multiplicative inverse modulo $n$. It therefore permutes the reduced residue system, givingThe product is also a unit modulo n, so it can be cancelled. This proves the Fermat-Euler theorem. For $n=1$ every modular congruence is automatic; one may use $\phi (1)=1$.

For a prime number $p$, the Euler totient function has value $\phi (p)=p-1$. Thus $a^{p-1}\equiv 1(\mathrm{mod}p)$ whenever $p\nmid a$. Multiplying by $a$, and treating $p\mid a$ separately, gives the all-integer version of Fermat's little theorem:Finally, Wilson theorem states that for every prime number $p$, $\boxed{(p-1)!\equiv -1(\mathrm{mod}p)}$. Its converse is also true: an integer $n>1$ is prime exactly when $(n-1)!\equiv -1(\mathrm{mod}n)$. For the converse, any proper divisor $1<d<n$ would divide both $(n-1)!$ and $n$, contradicting that congruence.

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.

Measure the second register of $|f⟩$. If the result is $f(x_0)$, injectivity within one period collapses the first register to the periodic coset stateApply ${\mathrm{QFT}}_N$ and measure the first register. By the stated transform identity, the outcome isfor a uniformly random $l\in \{0,\dots ,r-1\}$.

Reduce the rational numberto lowest terms and call its denominator $q$. The exact period recovery from a Fourier sample givesThus $q=r$ exactly when $l$ is coprime to $r$.

The success can be certified with two classical function evaluations. Compute $f(0)$ and $f(q)$. Since $q\mid r$ and $0\le q\le r$, periodicity and injectivity within a period implyReport $q$ only when this equality holds; otherwise report failure. The success probability isusing the standard lower bound for the Euler totient function. This is heralded exact quantum period finding when the period divides the register size.

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are

The prime-factor formula for the Euler totient function isLetThe factors with $p\le y$ contribute at mostby the Mertens third theorem. For $p>y$, the number of distinct prime divisors of $n$ is at most $\log n/\log y$, and henceThus the large-prime product is $1+o(1)$, andTaking reciprocals gives, uniformly as $n\to \infty$,

Let $L$ be a splitting field of $f(X)=X^m-1$ over $k$. Its formal derivative isThe assumption on the characteristic of $k$ says that $m\ne 0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f^{\prime }(\alpha )=m{\alpha }^{m-1}\ne 0$. Thus $f$ and $f^{\prime }$ have no common root, so $f$ is a separable polynomial. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the separability of roots of unity.

Fix a primitive root of unity ${\zeta }_m=e^{2\pi i/m}$. Define the $m$th cyclotomic polynomial byIt is monic and has degree given by the Euler totient function $\phi (m)$. Partitioning all $m$th roots of unity according to their exact multiplicative order gives the cyclotomic factorizationWe now prove by mathematical induction that ${\Phi }_m\in \mathbb{Z}[X]$. The base case is ${\Phi }_1=X-1$. If the result holds for every proper divisor $d$ of $m$, thenis monic and belongs to $\mathbb{Z}[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb{Z}[X]$. The factorization over $\mathbb{C}$ says that the remainder is zero and the quotient is ${\Phi }_m$, proving the claim.

It remains to prove irreducibility of cyclotomic polynomials. Let $f\in \mathbb{Z}[X]$ be a monic irreducible polynomial dividing ${\Phi }_m$, let $\epsilon$ be a root of $f$, and write ${\Phi }_m=fg$. We claim that ${\epsilon }^p$ is a root of $f$ for every prime number $p\nmid m$. Otherwise ${\epsilon }^p$ is a root of $g$, so $\epsilon$ is a root of $g(X^p)$. Since $f$ is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$, Gauss lemma for polynomials givesAfter reduction of an integer polynomial modulo a prime, the Frobenius endomorphism givesChoose an irreducible factor $h$ of the nonconstant monic polynomial $\bar{f}$. Then $h\mid \bar{g}$, so $h^2\mid \bar{f}\bar{g}={\bar{\Phi }}_m$. The cyclotomic factorization would then make $h^2$ divide $X^m-1$ over ${\mathbb{F}}_p$. This is impossible because its derivative $m{X}^{m-1}$ is coprime to it when $p\nmid m$.

Thus $f({\epsilon }^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f({\epsilon }^a)=0$ whenever $(a,m)=1$. These are all $\phi (m)$ primitive $m$th roots, soBecause $f$ divides ${\Phi }_m$, equality holds and $f={\Phi }_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb{Q}$.

Finally, direct multiplication givesAlsoApplying the cyclotomic factorization to the two factors on the right givesand thereforeSince $X^2-X+1={\Phi }_6(X)$, cancellation yieldsThe thirtieth cyclotomic polynomial is irreducible over $\mathbb{Q}$ by the theorem just proved, so $F$ is irreducible.

The Euler totient function $\varphi (n)$ is the number of residue classes modulo $n$ that are coprime to $n$. The Chinese remainder theorem says that for pairwise coprime moduli $m_i$, reduction induces a bijectionIt restricts to a bijection of unit groups. If $n=p_1\cdots p_r$ is squarefree, thenEach of the primes $3,5,17,257$ is one more than a power of two, so every squarefree product of them has power-of-two totient. Ten odd examples are

If $m$ and $n$ are coprime, every divisor of $mn$ is uniquely $ab$ with $a\mid m$ and $b\mid n$. Hence, for a multiplicative arithmetic function $f$,

The Möbius function is $\mu (n)=0$ if a prime square divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct primes. The Euler totient function $\varphi (n)$ counts residues modulo $n$ coprime to $n$. From the prime factorizations,Both sides of the second identity are multiplicative, and at a prime power $p^a$,Therefore

Start two registers in $|0⟩|0⟩$. Apply ${\mathrm{QFT}}_N$ to the first and then reversibly evaluate $f$ into the second, obtainingThe evaluation can be implemented by repeated squaring and reversible modular multiplication using a number of elementary arithmetic operations polynomial in $\log N$.

Measure the second register. Because $f$ is injective within one period and $r\mid N$, the first register becomes a uniformly weighted periodic cosetThe quantum Fourier transform of a periodic coset state followed by measurement returnsReduce the rational number $c/N$ to lowest terms. Sinceits denominator is $q=r/\gcd (s,r)$. Test whether $a^q\equiv 1(\mathrm{mod}N)$ by repeated squaring. The promised injectivity means that $r$ is the least positive period, and $q\mid r$, so this congruence holds exactly when $q=r$. The test therefore certifies whether the run succeeded; this is heralded exact quantum period finding when the period divides the register size.

One run succeeds with probabilitybecause success is equivalent to $s$ being coprime to $r$. A standard estimate for the Euler totient function gives $\phi (r)/r\ge C/\log \log r$ for all sufficiently large $r$, with the finitely many smaller cases absorbed by changing the positive constant $C$. Repeating independently $O(\log \log N)$ times therefore makes the probability that every run fails at most $1/2$. Every step, including the repetitions and the classical verification, takes time polynomial in $\log N$.

For a positive modulus $n$, a reduced residue system contains exactly one representative of each residue class coprime to $n$. Its size is the Euler totient function $\phi (n)$. Multiplication by any unit modulo n permutes these classes, which proves the Fermat-Euler theorem after multiplying and cancelling their unit product.

For $n\ge 3$, this explicit lower bound on the Euler totient function implies $\phi (n)/n=\Omega (1/\log \log n)$ as $n\to \infty$. Here ${\gamma }_E$ is the Euler--Mascheroni constant. The prime-product formula also gives $\phi (r)/r\ge \phi (N)/N$ whenever $r\mid N$, making the bound useful for certified quantum period finding. The relevant statement is a lower bound: the claimed universal upper estimate $\phi (n)=O(n/\log \log n)$ would fail at primes. Equations (3.41)–(3.42) of Theorem 15 provide the primary bound: denisevellachemla.eu/Rosser-Schoenfeld-1962.pdf .

A semiprime is a composite number that is a product of exactly two prime numbers, counted with multiplicity. For example, $6=2\cdot 3$ and $9=3^2$ are semiprimes. Distinct factors give $\phi (pq)=(p-1)(q-1)$ for the Euler totient function, whereas equal factors give $\phi (p^2)=p(p-1)$. This distinction matters when constructing a uniform coprime state for a semiprime.