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The Dirichlet convolution of arithmetic functions $f,g$ is the arithmetic functionIn the first sum, $d$ runs over the positive integer divisors of $n$; the second uses ordered pairs of positive integers whose product is $n$. Both sums are finite. Each pair contributes one weight from $f$ and one from $g$, so Dirichlet convolution combines information attached to complementary factors. It turns a divisor sum into an operation on arithmetic functions. ^[ref]

For $n=12$ the six ordered pairs are $(1,12),(2,6),(3,4),(4,3),(6,2),(12,1)$. The constant-one arithmetic function gives one per pair, while the identity arithmetic function weights the first factor by its size:Here $\tau$ is the divisor function and ${\sigma }_1$ is the sum-of-divisors function. By comparison, the pointwise product of arithmetic functions gives $(1\cdot 1)(12)=1$ and $(\mathrm{id}\cdot 1)(12)=12$.

Commutativity follows by swapping the two entries of every ordered pair. Associativity follows by reindexing either grouping over the same ordered tuples of three positive integers:For example, $((1\ast 1)\ast 1)(12)$ counts ordered tuples of three positive integers with product $12$. Distribute the two occurrences of the prime factor $2$ among three entries in $\left(\genfrac{}{}{0ex}{}{2+2}{2}\right)=6$ ways, and the one occurrence of the prime factor $3$ in $\left(\genfrac{}{}{0ex}{}{1+2}{2}\right)=3$ ways. These binomial coefficients are the stars and bars counts of exponent choices in the nonnegative integers; prime factorization therefore gives $6\cdot 3=18$ such tuples. Directly, the same Dirichlet convolution is ${\sum }_{d\mid 12}\tau (d)=1+2+2+3+4+6=18$.

The Identity for Dirichlet convolution is $\epsilon$, and the Möbius function reverses a divisor sum because $\mu \ast 1=\epsilon$. This makes Dirichlet convolution useful both for counting factorizations and for recovering an arithmetic function from its divisor sum.

The Identity for Dirichlet convolution is the arithmetic functionFor every arithmetic function $f$, it satisfies $f\ast \epsilon =\epsilon \ast f=f$. Indeed, in $(f\ast \epsilon )(n)={\sum }_{d\mid n}f(d)\epsilon (n/d)$, only $d=n$ survives. For example,At $n=1$ the sole term is $f(1)\epsilon (1)=f(1)$. This gives a useful way to check Dirichlet convolution calculations: inserting $\epsilon$ leaves every value unchanged. The three arithmetic functions $1$, $\mathrm{id}$, and $\epsilon$ all take value $1$ at $1$, but at $12$ the constant-one arithmetic function, the identity arithmetic function, and the Identity for Dirichlet convolution take values $1$, $12$, and $0$, respectively. ^[ref]

The Möbius function is the inverse of the constant-one arithmetic function under Dirichlet convolution, meaning $\mu \ast 1=1\ast \mu =\epsilon$. The Möbius divisor-sum identity gives a short proof: for $n>1$ the nonzero terms expand ${\prod }_{p\mid n}(1-1)=0$, while at $1$ the sum is $\mu (1)=1$. At $12$, the computation isThe zeros occur because $4$ and $12$ are divisible by the square of the prime number $2$. Consequently, if $F=f\ast 1$ is a divisor sum, then associativity gives $F\ast \mu =f\ast (1\ast \mu )=f\ast \epsilon =f$: this is Möbius inversion.

The Mertens function is the summatory Möbius function

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.

The Möbius function assigns a cancellation weight to each positive integer:Thus a squarefree integer contributes a sign determined by its number of distinct prime factors; a repeated prime factor makes the weight zero. This packages the alternating signs of inclusion-exclusion into an arithmetic function. ^[ref]

For example, $\mu (2)=\mu (3)=-1$, $\mu (6)=1$, and $\mu (30)=(-1{)}^3=-1$, since $30=2\cdot 3\cdot 5$. In contrast, $\mu (4)=\mu (12)=\mu (60)=0$: all three have the square of the prime number $2$ as a divisor. On prime powers, $\mu (p)=-1$ and $\mu (p^a)=0$ for $a\ge 2$.

The Möbius function is a multiplicative function: for coprime integers, their prime factors are disjoint, so either a squared prime factor forces both sides of $\mu (mn)=\mu (m)\mu (n)$ to vanish, or the counts of distinct prime factors add and their signs multiply. It is not a completely multiplicative function, since $\mu (4)=0\ne 1=\mu (2{)}^2$. Its Möbius divisor-sum identity makes it the inverse of the constant-one arithmetic function under Dirichlet convolution: $\mu \ast 1=\epsilon$, the Identity for Dirichlet convolution. This is why the small set of weights $-1,0,1$ can undo an entire divisor sum.

Möbius inversion recovers an arithmetic function from its divisor sum. For arithmetic functions $f,g$, the following statements are equivalent, with each equality required for every positive integer $n$:Here $d$ ranges over positive integer divisors of $n$ and $\mu$ is the Möbius function. No multiplicativity assumption is needed, and both sums are finite. The right side can also be written ${\sum }_{d\mid n}\mu (n/d)g(d)$ by exchanging complementary integer divisors. ^[ref]

The divisor sum $g(n)$ bundles $f(n)$ together with values at smaller integer divisors. Möbius inversion subtracts aggregates indexed by one distinct prime factor, adds those indexed by two, and continues with the alternating signs of inclusion-exclusion. A divisor containing a squared prime factor receives weight zero. The Möbius divisor-sum identity ensures that every unwanted contribution cancels exactly.

In Dirichlet convolution notation, $g=f\ast 1$ and $\mu \ast 1=\epsilon$, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. Associativity and commutativity give the short proof:Convolving $f=\mu \ast g$ with $1$ proves the reverse implication. At $n=1$, the formula reduces to $f(1)=g(1)$ because $\mu (1)=1$.

For a worked example, take $f(n)=\phi (n)$, Euler's totient function. The totient divisor-sum identity says that $g(n)=n$, the identity arithmetic function:This identity comes from classifying the integers $1,\dots ,n$ by their greatest common divisor with $n$: the group with $\gcd (j,n)=h$ has $\phi (n/h)$ members, since $j/h$ and $n/h$ are coprime integers. Summing the group sizes gives $n$, as detailed under totient divisor-sum identity.

Now undo this aggregation with the Möbius function:The nonzero terms use $d=1,2,3,6$; $d=4,12$ vanish because they contain the square of the prime number $2$. The four surviving integers coprime to $12$ are $1,5,7,11$.

This example also reveals why Möbius inversion is useful: it turns an easy count of all objects into a count satisfying a coprimality restriction. Expanding over subsets of distinct prime factors givesThe product is inclusion-exclusion over multiples of the prime factors. For $12$, it gives $12(1-1/2)(1-1/3)=4$. More generally, whenever a counting problem produces a divisor sum, Möbius inversion lets that aggregate be reversed without solving a new system of equations.

A number-theoretic sieve processes many integers together by marking or updating those sharing a divisor, rather than testing each integer independently. The sieve of Eratosthenes marks composite multiples of successive prime numbers; related sieves accumulate a smallest prime factor, Euler's totient function, or Möbius function for every argument up to a bound. The useful structure is that the multiples of a fixed integer form an arithmetic progression, so one loop can update all affected entries in an array.

For example, initializing an array to true and then marking $2p,3p,\dots$ for each surviving prime number $p$ leaves precisely the prime numbers unmarked, apart from the separately excluded zero and one. Every composite number has a prime factor, which proves that every composite is eventually removed. Starting the marking at $p^2$ avoids work already performed by smaller prime numbers. Different sieves use different update rules and time complexity bounds; the name alone does not imply the complexity of the sieve of Eratosthenes.

The Möbius function satisfies $\mu (p^e{)}^2=1$ for $e=0,1$ and is zero for $e\ge 2$. ThereforeThe inverse local factor is $(1+p^{-s}{)}^{-1}={\sum }_{e\ge 0}(-1{)}^e{p}^{-es}$. Consequently the inverse under Dirichlet convolution is the Liouville functionwhere $\Omega (n)$ counts prime factors with multiplicity. Both functions are multiplicative functions. At $p^e$ with $e\ge 1$, their convolution is $(-1{)}^e+(-1{)}^{e-1}=0$, while its value at one is one. Multiplicativity then proves $h_1\ast h_2=I$ on every positive integer. Its Dirichlet series is $H_2(s)=\zeta (2s)/\zeta (s)$ for $\mathrm{Re}s>1$.

The equation ${\sigma }^{n}x=x$ says $x_{r+n}=x_r$ for every $r\in \mathbb{Z}$. Thus a solution is determined uniquely by the block $(x_0,\dots ,x_{n-1})$, repeated in both directions. It belongs to the subshift of finite type exactly when all transitions inside the block, and the closing transition from $x_{n-1}$ to $x_0$, are allowed. Hence these fixed points of the $n$th iteration of a map are in bijection with the closed transition chains of $n$ steps with a specified symbol at time zero. By the counting argument in part (b), the number starting at $i$ is $(A^n{)}_{ii}$. Summing over $i$ proves the trace formula for periodic points of a subshift:Every closed transition chain extends by periodic repetition, so the extension issue in part (b) does not affect this formula. We count points, not periodic orbits: a least-period-$d$ periodic orbit contributes its $d$ distinct points.

Here period $n$ means fixed by ${\sigma }^n$, so the least period may divide $n$. If $P_d$ denotes the number of points of least period exactly $d$, division with remainder shows that a point is fixed by ${\sigma }^n$ exactly when its least period divides $n$. Consequentlywhere ${\mu }_{\mathrm{M}}$ is the Möbius function and the second equality is Möbius inversion. For example, $A=(1)$ has one point fixed by every power but no point of least period two; interpreting the printed formula as an exact-period count would therefore be false.

Define the Möbius function by $\mu (1)=1$, $\mu (n)=(-1{)}^r$ for a product of $r$ distinct primes, and $\mu (n)=0$ if a prime square divides $n$. For $n>1$, summing over square-free divisors gives ${\sum }_{d\mid n}\mu (d)=(1-1{)}^r=0$, while the sum is one at $n=1$.

Both Dirichlet series $\sum n^{-s}$ and $\sum \mu (n)n^{-s}$ converge absolutely for $\sigma =\mathrm{Re}s>1$. Their product may therefore be regrouped by the product of the indices:This simultaneously proves $\boxed{\zeta (s)\ne 0\text{ for }\sigma >1}$ andIn particular $|1/\zeta (\sigma +it)|\le \sum |\mu (n)|n^{-\sigma }\le \zeta (\sigma )$.

Under the assumed analytic continuation, $\zeta$ is holomorphic near $s_0=1+i{t}_0$ when $t_0\ne 0$, and $\zeta (\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$. If $s_0$ were a zero of order $m\ge 2$, local factorization would give $|1/\zeta (s_0+h)|\ge c{h}^{-m}$ for sufficiently small positive $h$. The preceding absolutely convergent bound gives instead $|1/\zeta (1+h+i{t}_0)|=O(h^{-1})$, a contradiction. Thus $\boxed{m\le 1}$. This reciprocal Dirichlet-series bound on boundary-zero multiplicity does not assert that any such boundary zero exists.

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

Let $\mu$ be the Möbius function. Its Möbius divisor-sum identity isIndeed, for $n>1$ the sum is the expansion of ${\prod }_{p\mid n}(1-1)$; nonsquarefree divisors contribute zero. Thus, for arithmetic functions $a,b$,To prove Möbius inversion, substitute the first formula into the second and collect the coefficient of $a(e)$. It is ${\sum }_{d\mid n/e}\mu (d)$, which is one for $e=n$ and zero otherwise. Conversely the same divisor interchange recovers $b$ from the displayed formula for $a$. In Dirichlet convolution notation this is simply $\mu \ast 1=\epsilon$, where $\epsilon$ is supported at one.

The Prime number theorem with classical zero-free-region error states that some absolute $c>0$ satisfiesHere $\Lambda$ is the Von Mangoldt function. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi (x)=\mathrm{Li}(x)+O(x{e}^{-c\sqrt{\log x}})$. The contributions of proper prime powers are $O(\sqrt{x}{\log }^{2}x)$ and can be absorbed into this error. We use $\Psi$ to keep it distinct from the Fourier transform appearing in Question 5.

Here is a direct deduction of the requested Mertens bound from a log-integrable Chebyshev error from the stated Prime number theorem. First, the divisor identity givesSecond, the exact Dirichlet convolution identityfollows from $\log =1\ast \Lambda$: multiplication of a Dirichlet convolution by $\log n$ differentiates its two factors, so applying it to $\mu \ast 1=\epsilon$ and convolving again with $\mu$ gives this formula. Summing it yieldsThe last harmonic sum is $O(1)$. To see this without an endpoint approximation, split $x/d$ into dyadic ranges $[2^j,2^{j+1})$. In each range $\sum 1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log 2}}$. Their sum over $j\ge 0$ converges. The finitely many terms with $1\le x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. ThereforeFor the Mertens function $M(x)={\sum }_{n\le x}\mu (n)$,The second sum has absolute value at most ${\sum }_{n\le x}\log (x/n)=O(x)$, by an integral comparison or the Stirling formula. ConsequentlyThe absolute value in this conclusion is present in the original PDF.

The Möbius function has $\mu (1)=1$, vanishes on integers divisible by a square of a prime, and equals $(-1{)}^r$ on a product of $r$ distinct primes. Factoring the divisor sum prime by prime givesThis is the Möbius divisor-sum identity.

For $T/2\le t\le T$ and $0<\sigma \le 1$, the Hardy-Littlewood approximation to the Riemann zeta function at cutoff $T$ givesIndeed $T\ge |t|/\pi$, and the omitted integral term has size at most $T^{1-\sigma }/|s-1|\ll T^{-\sigma }$. Multiply by ${\sum }_{m\le M}\mu (m)m^{-s}$. Its absolute value is at mostwhere the elementary inequality follows from $m^{1-\sigma }\le M^{1-\sigma }$. Reindexing the finite double sum yields coefficients $a_n={\sum }_{m\mid n,m\le M,n/m\le T}\mu (m)$, with no terms for $n>MT$. For $n\le \min (M,T)$ all divisors meet both restrictions, so the Möbius divisor-sum identity gives $a_1=1$ and $a_n=0$ for $1<n\le \min (M,T)$. Hence the truncated Möbius inverse identity for the Riemann zeta function isThe displayed error is uniform in $0<\sigma \le 1$ and the stated height interval.

Finite rearrangement and the Von Mangoldt divisor identity giveFor $m=\lfloor x\rfloor$, integral comparison of the increasing logarithm gives $\log (m!)=m\log m-m+O(\log (m+1))$. Moving $m$ to real $x$ changes the main term by $O(\log (x+1))$. Thus the integrated Chebyshev sum hasFor $x\ge 2$, chooseThese are logarithmic Möbius function weights; they need not be upper-bound sieve weights. For $n\le x$, the Möbius inversion identities $\mu \ast 1=ϵ$ and $\mu \ast \log =\Lambda$ giveHence the expression suggested in the question isWe will use the elementary Möbius harmonic logarithmic moments, $A_j(x)={\sum }_{d\le x}\mu (d){\log }^j(x/d)/d$, for whichHere is a proof of the needed estimates. From ${\sum }_d\mu (d)\lfloor x/d\rfloor =1$, replacing floors by $x/d+O(1)$ gives $A_0=O(1)$. Also ${\sum }_d\mu (d)H_{\lfloor x/d\rfloor }/d=1$, because $\mu \ast 1=ϵ$. Substituting the permitted harmonic number estimate $H_{\lfloor t\rfloor }=\log t+\gamma +O(1/t)$ gives $A_1+\gamma A_0=1+O(1)$, so $A_1=O(1)$.

For $A_2$, the weighted Dirichlet hyperbola method gives the harmonic divisor sum expansionExplicitly, for $r=\lfloor \sqrt{t}\rfloor$, $B(t)=2{\sum }_{a\le r}{H}_{\lfloor t/a\rfloor }/a-H_r^2$. Substitute the harmonic estimate and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$, obtained by unit-interval integral comparison. This proves the expansion without a prime-distribution theorem. Since $\tau =1\ast 1$, we have $\mu \ast \tau =1$, and thereforeThe summed error is bounded by $x^{-1/2}{\sum }_{d\le x}{d}^{-1/2}\log (2x/d)=O(1)$, by integral comparison. Thus $A_2=2\log x+O(1)$.

Finally substitute the asymptotic for $T$ into $F$. The main term isThe error sum is $O(x)$: partition the integers $d$ into $x/2^{j+1}<d\le x/2^j$; each interval contributes $O(x(j+1{)}^2/2^j)$, and the resulting series converges. Using the moment estimates proves the Selberg symmetry formula:In particular, the factor two comes from the quadratic logarithmic moment, rather than from an assumption of the Prime number theorem.

The Euler totient function counts integers $1\le a\le n$ with $\gcd (a,n)=1$. The Möbius function has $\mu (1)=1$, is zero when a prime square divides $n$, and equals $(-1{)}^r$ when $n$ is a product of $r$ distinct primes. Consequentlybecause for $n>1$ the sum over squarefree divisors is $(1-1{)}^r$.

The Möbius inversion formula is $g(n)={\sum }_{d\mid n}\mu (d)f(n/d)$. Substituting the divisor sum defining $f$ and collecting the coefficient of $g(e)$ givesThis proof uses only finite sums, so no convergence hypothesis is needed.

Classifying $1,\dots ,n$ by their greatest common divisor with $n$ gives ${\sum }_{d\mid n}\phi (d)=n$: the class with greatest common divisor $n/d$ contains exactly $\phi (d)$ integers. Applying the Möbius inversion formula yieldsThe Riemann zeta function is $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\mathrm{Re}s>1$. Absolute convergence justifies multiplication of Dirichlet series. The divisor identity for the Möbius function gives $\sum \mu (n)n^{-s}=1/\zeta (s)$ there. For $\mathrm{Re}s>2$ both factors below converge absolutely, and their Dirichlet convolution is the Euler totient function:

The Möbius function has $\mu (1)=1$, $\mu (n)=0$ when $n$ is divisible by a prime square, and $\mu (n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is initially the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\sigma =\Re s>1$. Both this Dirichlet series and $\sum \mu (n)n^{-s}$ have absolute convergence, because $|\mu (n)|\le 1$ and $\sum n^{-\sigma }<\infty$.

For a finite set of primes $p\le y$, multiplication of absolutely convergent geometric series and unique prime factorization giveEvery omitted integer has a prime factor exceeding $y$, so it exceeds $y$. The difference from $\zeta (s)$ has absolute value at most ${\sum }_{n>y}{n}^{-\sigma }\to 0$. This proves the Euler product, with the product interpreted as the limit of its finite products:

For the reciprocal identity, absolute convergence justifies multiplying and regrouping the two Dirichlet series by their product index. Their Dirichlet convolution coefficient at $m$ isThe coefficient calculation uses only the squarefree integers dividing $m$. Consequently the reciprocal identity is an identity of absolutely convergent series:

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

An arithmetic function $f$ is multiplicative when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the Dirichlet convolution of multiplicative functions satisfiesand $(f\star g)(1)=1$, so it is multiplicative.

Let $\epsilon$ be the Identity for Dirichlet convolution, with $\epsilon (1)=1$ and $\epsilon (n)=0$ for $n>1$. The Möbius function satisfies $\mu \star 1=\epsilon$. Therefore, using commutativity and associativity,This is the Möbius inversion formula.

The Möbius function is defined by $\mu (1)=1$ and, for $n>1$,It is a multiplicative arithmetic function: $\mu (mn)=\mu (m)\mu (n)$ whenever $m$ and $n$ are coprime.

The summand $\mu (d{)}^2/\varphi (d)$ is multiplicative, so its divisor sum is multiplicative. For a prime power $p^a$ with $a\ge 1$, only $d=1,p$ contribute andMultiplying these local identities over the prime divisors of $n$ gives

For the final claim, choose distinct primes $p_0,\dots ,p_k$. The moduli $p_j^2$ are pairwise coprime, so the Chinese remainder theorem gives an integer $a$ satisfyingEvery positive integer $n\equiv a(\mathrm{mod}{\prod }_j{p}_j^2)$ then has $p_j^2\mid n+j$. HenceThere are infinitely many positive representatives of this congruence class.