The symmetric form of the functional equation of the Riemann zeta function isThe complex conjugation identity $\zeta (\overline{s})=\overline{\zeta (s)}$ and the functional equation show that every Nontrivial zero of the Riemann zeta function $\rho$ is accompanied by $\overline{\rho }$, $1-\rho$, and $1-\overline{\rho }$.

Let $\rho =\beta +i\gamma$ be the given nonreal zero. If $\beta >1/2$, use $\rho$ itself; if $\beta <1/2$, use $1-\rho$, whose real part is $1-\beta >1/2$. The resulting zero cannot have real part greater than one, by the stated zero-free half-plane. It therefore has real part in $(1/2,1]$.

If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, expanding the product over a finite set of primes and using unique prime factorization gives the sum over integers having no other prime factors. Moreover,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

For each prime number $p\nmid a$, the congruence $an+1\equiv 0(\mathrm{mod}p)$ removes exactly one residue class of $n$ modulo $p$; for $p\mid a$, it removes none. The dimension-one upper-bound sieve therefore givesSeparating the primes that divide $a$ bounds the product byThe ratio form of Mertens theorem says that the first product is $\ll \log w/\log z$. Hence

The Dirichlet hyperbola method counts each factorization $n=ab$ once and gives, with $y=\lfloor \sqrt{x}\rfloor$,Using $\lfloor x/a\rfloor =x/a+O(1)$ and the harmonic number estimate $H_y=\log y+\gamma +O(1/y)$, where $\gamma$ is the Euler--Mascheroni constant, we obtainThus one may take $C=2\gamma -1$.

Let $1(n)=1$. The basic Von Mangoldt divisor identity isbecause if $n={\prod }_p{p}^{v_p(n)}$, the right-hand side is ${\sum }_p{v}_p(n)\log p=\log n$. In terms of Dirichlet convolution, this says $\log =1\ast \Lambda$. Since the Möbius function is the convolution inverse of $1$, convolving with $\mu$ gives $\Lambda =\mu \ast \log$. Consequently

Write each hyperplane as $h_j(x)=0$, normalized so that $h_j(0)=1$, and putThen $P(0)=1$ and $P$ vanishes at every other point of ${\mathbb{F}}_q^n$. Reduce $P$ modulo $x_i^q-x_i$ in every variable. This preserves its function on ${\mathbb{F}}_q^n$, does not increase total degree, and gives the unique representative with each variable degree at most $q-1$.

The unique reduced polynomial for the delta function at zero iswhose total degree is $(q-1)n$. HenceUniqueness follows equally from the Alon-Tarsi lemma on the grids $A_i={\mathbb{F}}_q$.

The Alon-Tarsi lemma says that if $\mathrm{deg}f\le d_1+\cdots +d_n$ and $A_i$ consists of $d_i+1$ distinct field elements, thenThis follows by applying univariate Lagrange interpolation successively in each variable.

If the displayed coefficient is nonzero, at least one summand has $f(a_1,\dots ,a_n)\ne 0$. This is the coefficient form of the Combinatorial Nullstellensatz.

The required prime-power form of the Frankl-Wilson theorem is: if $A_1,\dots ,A_m\subseteq [n]$, no $|A_i|$ is divisible by $p^r$, and every intersection of $k$ distinct members has size divisible by $p^r$, thenIt follows by associating to each $A_i$ its incidence vector augmented by a constant coordinate and applying the Frankl-Wilson polynomial independence lemma to the $(k-1)$ layers of multilinear intersection polynomials. The hypotheses make the diagonal evaluations nonzero modulo $p^r$ and every $k$-fold off-diagonal evaluation zero; independence leaves at most $(k-1)(n+1)$ polynomials. Applying the theorem gives the desired bound.

When $r=1$, the argument works directly over ${\mathbb{F}}_p$ without the constant-coordinate lift and yields the stronger boundTherefore a family of size $(k-1)(n+1)$ does not exist.

Let $v_i\in {\mathbb{F}}_2^n$ be the incidence vector of $S_i$. Thenfor $i\ne j$. If ${\sum }_i{c}_i{v}_i=0$, taking the dot product with $v_j$ gives $c_j=0$. The vectors are linearly independent, so $m\le n$.

Since $D(x)=xC(x)=(1-\sqrt{1-4x})/2$,Rationalizing and using the Catalan generating function givesThus the coefficient of every $t^k$, $0\le k\le n$, is $C_n$. The number of bridges with exactly $2k$ negative steps is consequently independent of $k$.

A strictly negative primitive excursion of semilength $n$ has all $2n$ steps negative, so weight ${\sqrt{t}}^{2n}=t^n$. Reflection in the axis identifies it with a strictly positive excursion, givingEvery bridge has a unique decomposition at successive returns to the axis into positive or negative primitive excursions. The sequence construction therefore has generating functionIts exponent of $t$ is half the number of negative steps, proving the asserted interpretation of $f(t,n)$.

Deleting the first up-step and last down-step of a strictly positive walk of semilength $n$ and lowering the remainder by one gives a Dyck path of semilength $n-1$, bijectively. Hence $D_n=C_{n-1}$ and

A nonempty Dyck path decomposes uniquely as an up-step, a Dyck path, a down-step, and another Dyck path. Marking each matched outer pair by $x$ givesThe solution with constant term one is

A proper colouring using exactly $k$ colours first partitions the vertices into $k$ nonempty independent colour classes and then injectively assigns $k$ of the $x$ named colours to those classes. These choices numberSumming over $k$ counts every proper colouring exactly once; the expression is the chromatic polynomial ${\chi }_G(x)$.

In an independent-block partition of $P_n$, either $v_1,v_n$ lie in different blocks, giving a partition valid for $C_n$, or they lie in the same block. Contracting those endpoints in the second case gives an independent-block partition of $C_{n-1}$. This bijection proves the recurrence. Multiplying by $x^{\underline{k}}$ and summing givesUsing ${\chi }_{P_n}=x(x-1{)}^{n-1}$ and induction from $C_2=P_2$ yields

The sum is the chromatic polynomial of the path, so

Partitions of the vertices into independent blocks are the unlabeled colour-class partitions counted by the Graphical Stirling number. ThereforeSince ${\chi }_{P_n}(x)=x(x-1{)}^{n-1}$, the ordinary Stirling identity applied to $x-1$ givesUniqueness in the falling-factorial basis proves the claim.

For a positive integer $x$, count functions $[n]\to [x]$ by the number $k$ of nonempty fibres. Their fibres form a $k$-block partition in $\left\{\begin{array}{c}n\\ k\end{array}\right\}$ ways, and the blocks receive distinct images in $x^{\underline{k}}$ ways. HenceBoth sides are polynomials of degree $n$ agreeing at every positive integer, so this is a polynomial identity.

The Stirling number of the second kind $\left\{\begin{array}{c}n\\ k\end{array}\right\}$ counts partitions of an $n$-element set into $k$ nonempty unlabeled blocks. Distinguishing the block containing the last element gives

Give every edge weight one and delete the row and column for $v_n$ from the out-Laplacian. Expanding the resulting banded determinant along its last available row givesThe characteristic roots are $2$ and $-1$, and the initial values giveBy the directed matrix-tree theorem, this determinant is exactly the number of directed spanning trees rooted towards $v_n$.