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Let $⟨x⟩$ denote the distance from $x$ to the nearest prime power. For $x\ge 2$ not an integer and $T\ge 2$, the truncated Riemann–von Mangoldt explicit formula isZeros are counted with multiplicity. The constant term comes from $s=0$, the logarithm collects the trivial zeros $-2,-4,\dots$, and the finite sum contains the nontrivial zeros. Enlarging the implied constant covers $1\le T<2$.

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.

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.

Write $\chi ={\chi }^{\lambda }$. By the Hook-length formula,Suppose the order of a group element $g$ did not divide this product. Then for some prime number $p$, the highest prime power $p^a$ dividing $|g|$ would not divide the hook product. One cycle of $g$ has length divisible by $p^a$, whereas no hook length of $\lambda$ is divisible by $p^a$. In particular there is no removable rim hook having that cycle length. Applying the Murnaghan–Nakayama rule first to this cycle gives $\chi (g)=0$, a contradiction. This is the symmetric-group character co-degree vanishing criterion, and its contrapositive proves

For unit-modulus prime values, pretentious distance is the Euclidean metric on the finite vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$. On whole arithmetic functions it is only a pseudometric: it sees neither values on prime powers nor primes above the cutoff. Even for every cutoff, the functions one and ${\mu }^2$ have distance zero but differ on nonsquarefree integers. The norm representation proves the triangle inequality and identifies exactly the quotient on which separation holds.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

If $K/{\mathbb{Q}}_p$ is finite of degree $d$, then as a topological abelian groupwhere ${\mu }_{p^a}(K)$ is the finite group of all roots of unity in $K$ whose orders are prime powers dividing a power of $p$.