text:
\section{Introduction}
Let
\be \la{abc} y'=p(x)y^2+q(x)y^3\ee
be the Abel differential equation, where $x$ is real and $p(x)$ and $q(x)$ are continuous.
Equation \eqref{abc}
is said to have a center on a segment $[a,b]$ if all its solutions, with the initial
value $y(a)$ small enough, satisfy the condition $y(b)=y(a)$.

The problem of description of conditions implying a center for \eqref{abc}  is closely related with
the classical Poincar\'e center-focus problem about conditions implying that all solutions of the system
\be \la{sys1}
\left\{
\begin{array}{rcl}
\dot{x} & =& -y+F(x,y), \\
\dot{y} & =&x+G(x,y),\\
\end{array}
\right.
\ee
where $F(x,y),$ $G(x,y)$ are polynomials without constant and linear terms,
around zero are closed. Namely, it was shown in \cite{cher} that if $F(x,y),$ $G(x,y)$ are homogeneous polynomials of the same degree, then
one can construct trigonometric polynomials $f(\cos \phi,\sin \phi),$ $g(\cos \phi,\sin \phi)$ such that \eqref{sys1} has a center if and only if all
solutions of the equation
$$
\frac{d r}{d \phi}=f(\cos \phi,\sin \phi)\, r^2+g(\cos \phi,\sin \phi)\,
r^3
$$
with $r(0)$ small enough
are periodic on $[0,2 \pi]$.

Set
\be \label{form} P(x)=\int_0^{x}p(s)ds , \ \ \ Q(x)=\int_0^{x} q(s)ds.\ee
The following ``composition condition'' introduced in \cite{al} is  sufficient
for equation \eqref{abc} to have a center:
there exist $C^1$-functions $\tilde P,\tilde Q, W$ such that
\be \label{c} P(x)=\tilde P(W(x)), \ \ \ \ Q(x)=\tilde Q(W(x)),\ \ \ \ W(a)=W(b). \ee
Indeed, if \eqref{c} holds, then any solution of \eqref{abc} has the form $y(x)=\tilde y(W(x)),$ where
$\tilde y$ is a solution of the equation
$$ y'=\tilde P^{\prime}(x)y^3+\tilde Q^{\prime}(x)y^2,$$
implying that $y(a)=y(b)$, since $W(a)=W(b).$
\pagebreak

It is known that in general the composition condition is not necessary for \eqref{abc} to have a center (\cite{alw}). However,
it
is believed that, in the case where $p(x)$ and $q(x)$ are polynomials, equation \eqref{abc} has a center if and only if the
composition condition \eqref{c}
holds for some polynomials $\t P, \t Q, W\in \R[x]$
(see  \cite{bpy}, \cite{bry} for some  partial results in this direction).

In this paper we study the following ``parametric center problem''
for equation \eqref{abc}
with polynomial coefficients:
{\it under what conditions the equation
\be \la{abp} y'=p(x)y^2+\v q(x)y^3, \ \ \  p,q\in \R[x],
\ee
has a center for any $\v \in \R$?} Posed for the first time in the series of papers \cite{bfy2}, \cite{bfy3}, \cite{bfy4},  this problem turned out to be very constructive
and resulted in a whole area of new ideas and methods related to the so called ``polynomial moment problem''
(see the discussion below).
However, in its full generality the parametric center problem  remained unsolved (see the recent paper \cite{bpy1} for the state of the art), and the goal of this paper is to fill this gap.
Our main result is the following theorem.

\bt \la{t1} Abel differential equation \eqref{abp}
has a center on a segment $[a,b]$ for any $\v \in \R$ if and only if the antiderivatives $P=\int p(x) dx$ and $Q=\int q(x) dx$
satisfy  composition condition \eqref{c} for some polynomials $\t P,$ $\t Q$, $W$.
\et

The proof of Theorem \ref{t1} is based on the link between the parametric center problem and the ``polynomial moment problem''. Namely, it was shown in \cite{bfy4} that the parametric center implies the equalities
\be \la{mix}
\int_a^b P^id Q=0, \ \ \ i\geq 0, \ \ \  \int_a^b Q^id P=0, \ \ \ i \geq 0.
\ee
We will call the problem of description of solutions \eqref{mix} the ``mixed polynomial moment problem'', and the problem of
description of solutions
\be \la{mom}
\int_a^b P^id Q=0,  \ \ \ i\geq 0,
\ee
simply the ``polynomial moment problem''.

The polynomial moment problem
has been studied in many recent papers (see e.g. \cite{bfy2},\cite{bfy3} \cite{bfy4}, \cite{c}, \cite{acoun},
\cite{pa2}, \cite{pp}, \cite{pry}, \cite{mp1}, \cite{pak1},
\cite{pak}, \cite{ro}).  Again, the composition condition \eqref{c} is sufficient for equalities \eqref{mom} to be satisfied although in general is not necessary (\cite{acoun}). A complete solution of the polynomial moment problem was obtained in the recent papers \cite{mp1}, \cite{pak}. Namely, it was shown in \cite{mp1} that if polynomials $P,$  $Q$ satisfy \eqref{mom}, then there exist polynomials $Q_j$ such that $Q=\sum_j Q_j$ and
\be \l{cc}
P(x)=P_j(W_j(x)), \ \ \
Q_j(x)=V_j(W_j(x)), \ \ \ W_j(a)=W_j(b)
\ee
for some polynomials $P_j(z), V_j(z), W_j(z)$. Moreover, in
\cite{pak} polynomial solutions of \eqref{mom} were desc\-ribed in
an explicit form  (see Section 2 below).

In this paper we apply results of \cite{pak} to each of the two systems appeared in \eqref{mix} separately and show that the restrictions obtained
imply that any solution $P,Q$ of the mixed polynomial moment problem satisfy  composition condition \eqref{c}. Thus, in fact we prove the following ``moment''
counterpart of Theorem \ref{t1}.

\bt \la{t2} Polynomials $P, Q\in \R[x]$  satisfy equalities \eqref{mix} if and only if they
satisfy composition condition \eqref{c} for some polynomials $\t P, \t Q, W\in \R[x].$
\et

Although the center problem for the Abel equation with polynomial coefficients can be consi\-dered in the complex setting, in this paper we
work in the classical real framework. Thus,
we start the paper from the adaptation for the real case of the results of \cite{pak}, obtained over $\C$.
Namely, we show in Section 2 that possible ``types'' of solutions of the polynomial moment problem over $\R$ remain the same, although one of these types becomes ``smaller'' (Theorem \ref{mpr}).
Besides, in Section 2 we establish some important restrictions of the arithmetical nature on points $a,b$ for which there exist solutions of \eqref{mom}
which do not satisfy
the composition condition  (Corollary \ref{red}).

In Section 3 we apply  the results of Section 2 to \eqref{mix}, and prove Theorem \ref{t2}.
The main difficulties of the proof
stem from the fact that after solving systems in \eqref{mix} separately we arrive to
functional equations of the type
\be \la{eba} \sum_{j=1}^r V_j(W_j(x))=A(B(z)),\ee where $V_j,W_j,A,B$ are polynomials, and $r$ equals $2$ or 3. Such equations can be considered as generalizations of the functional equation
\be \la{ebaq} A(B(x))=C(D(x)),\ee studied by Ritt (\cite{rit}).
However, the well established methods for studying \eqref{ebaq}, related to the monodromy, cannot be applied to \eqref{eba} for $r>1$, and essentially the only method which remains is a  painstaking analysis of coefficients. Such an analysis in general leads to rather
complicated systems of equations, and Theorem \ref{t2} is deduced from restrictions on $P$ and $Q$ obtained from these systems combined with restrictions on possible values of $a$ and $b$.

\section{Polynomial moment problem over $\C$ and over $\R$}
\subsection{Solution of the polynomial moment problem over $\C$}
In this subsection we briefly recall a description of
$P,Q\in \C[z]$  satisfying \eqref{mom} for $a,b\in \C$. For more details we refer
the reader to \cite{pak}.

Recall that the Chebyshev polynomials (of the first kind) $T_n$ are defined by the formula $T_n(\cos \phi)=\cos(n \phi).$ It follows directly from this definition
that
\be \la{tyui} T_n(1)=1, \ \ \ \ T_n(-1)=(-1)^n, \ \ \ n\geq 0\ee
and
$$T_n\circ T_m=T_m\circ T_n=T_{mn},\ \ \ n,m\geq 1, $$
where the symbol $\circ$ denotes a composition of functions, $A\circ B=A(B(z))$.

An explicit expression for $T_n$ is
given by the formula
\be \la{cheb} T_n=\frac{n}{2}\sum_{k=0}^{[n/2]}(-1)^k\frac{(n-k-1)!}{k!(n-2k)!}(2x)^{n-2k},\ee
implying in particular that
\be \la{kots} T_n(-x)=(-1)^nT_n(x).\ee

Following \cite{pak}, we will call a solution $P, Q$ of \eqref{mom} reducible if composition condition \eqref{c} holds
for some $\t P, \t Q, W\in \C[x].$

\bt \l{mpc} (\cite{pak}) Let $P,$ $Q$ be non-constant complex polynomials
and $a,b$ distinct complex numbers such that equalities \eqref{mom} hold. Then, either $Q$ is a reducible solution of \eqref{mom}, or there exist
complex polynomials $P_j,$ $Q_j,$ $V_j,$ $W_j,$ $1\leq j \leq r,$
such that
$$Q=\sum_{j=1}^rQ_j, \ \ \
P=P_j\circ W_j, \ \ \
Q_j=V_j\circ W_j, \ \ \  \ \ \ W_j(a)=W_j(b).
$$
Moreover,
one of the following conditions holds:
\vskip 0.25cm
\noindent 1) $r=2$ and
$$P=U\circ z^{sn}R^n(z^n) \circ V,\ \ \ W_1=z^n \circ V, \ \ \ W_2=z^sR(z^n)\circ V,$$
where $R,$ $U$, $V$  are complex polynomials, $n>1$, $s> 0,$ $\GCD(s,n)=1;$

\vskip 0.25cm
\noindent 2)  $r=2$ and
$$P=U\circ T_{m_1m_2} \circ V,\ \ \ W_1= T_{m_1} \circ V, \ \ \
W_2= T_{m_2}\circ V,$$
where $U$, $V$ are complex polynomials, $m_1>1,$ $m_2>1$, $\GCD(m_1,m_2)=1;$

\pagebreak

\vskip 0.25cm
\noindent 3) $r=3$ and
$$P=U\circ z^{2}R^2(z^{2})\circ  T_{m_1m_2}
\circ V,$$
$$W_1=T_{2m_1} \circ V, \ \ \ W_2=T_{2m_2}\circ V, \ \ \
W_3= (zR(z^2) \circ T_{m_1m_2}) \circ V,$$
where $R,$ $U$, $V$ are complex polynomials,
$m_1>1,$ $m_2>1$ are odd, and $\GCD(m_1,m_2)=1.$
\et

We will call solutions 1), 2), 3) appearing in Theorem \ref{mpc} solutions of the first, the second, and the third type
correspondingly.

Notice that these sets of solutions are not disjointed. For example, if one of the parameters $n$, $m$ of a solution of the second type equals 2, then this solution is also
a solution of the first type. Indeed, if say $n=2,$ then $W_1=T_2\circ V=\mu\circ z^2\circ V,$ where $\mu =2x-1.$ On the other hand,  since $m$ is odd in view
of $\GCD(n,m)=1,$ the polynomial $W_2=T_m\circ V$  has the form $T_m=zR(z^2)\circ V$ by \eqref{cheb}.
Therefore,
$$P=(U\circ \mu)\circ z^{2}R^2(z^{2})\circ V,\ \ \  Q=((V_1 \circ \mu)\circ z^2+V_2\circ z R(z^2))\circ V,$$
and for the polynomials $\t W_1=z^2\circ V$ and $\t W_2=W_2=z R(z^2)\circ V$ the equalities
\be \la{aqwer} \t W_1(a)=\t W_1(b), \ \ \ \ \t W_2(a)=\t W_2(b)\ee hold.

Similarly, if the
parameters $a,b$ of a solution of the third type satisfy  $V(a)=-V(b)$, then  this solution is also a
solution of the first type. Indeed, $$V_1\circ T_{2m_1}+V_2\circ T_{2m_2}=\t V_1 \circ z^2$$ for some $\t V_1\in \C[z]$, while $$zR(z^2) \circ T_{m_1m_2}= z\t R(z^2)$$ for some $\t R\in \C[z]$, since $m_1,$ $m_2$  are odd. Therefore,
$$P=U\circ z^{2}\t R^2(z^{2})\circ V,\ \  \ Q=(\t V_1 \circ z^2+V_3\circ z\t R(z^2))\circ V,$$ and $\t W_1=z^2\circ V$ satisfies
$\t W_1(a)= \t W_1(b)$, since $V(a)=-V(b)$.

Finally, one can check that a solution of the third type is a solution of the second type if
$(T_{m_i}\circ V)(a)=(T_{m_i}\circ V)(b)$ for $i$ equals 1 or 2 (for more details concerning interrelations between different types of solutions see \cite{pak}, pp. 725-726).

\subsection{Lemmas related to $a,b$}
In this subsection we collect some results implying restrictions on the integration limits  $a,b$ appearing in solutions of the second and the third types.

\bl \la{skun} Let $T_{m_1},$ $T_{m_2},$ $T_{m_3}$ be the Chebyshev polynomials and $a$, $b$ be distinct complex numbers.
\vskip 0.1cm
\noindent a) Assume that
\be \la{volk} T_{m_1}(a)=T_{m_1}(b), \ \ \ \ \ \ T_{m_2}(a)=T_{m_2}(b).\ee
Then either  $T_l(a)=T_l(b)$ for $l=\GCD(m_{1},m_{2}),$ or \be \la{uraa} T_{m_1m_2}^{\prime }(a)=T_{m_1m_2}^{\prime }(b)=0.\ee
\vskip 0.1cm
\noindent b) Assume that \be \la{uuiioo} T_{m_1}(a)=T_{m_1}(b), \ \ \ \ \ \ T_{m_2}(a)=T_{m_2}(b), \ \ \ \ \ \ T_{m_3}(a)=T_{m_3}(b).\ee
Then there exists a pair of distinct indices $i_1,i_2,$ $1\leq i_1,i_2 \leq 3,$
such that  $T_l(a)=T_l(b)$ for $l=\GCD(m_{i_1},m_{i_2})$.

\el

\pr Choose $\alpha, \beta \in \C$ such that $\cos \alpha= a,$
$\cos \beta =b.$ Then equalities \eqref{volk}
imply the equalities
\be \la{100} m_1 \alpha = \v_1 m_1 \beta + 2\pi k_1, \ \ \
m_2 \alpha = \v_2 m_2 \beta+ 2\pi k_2, \ee where $\v_1=\pm 1,$ $\v_2=\pm 1,$
and $k_1,k_2\in \Z.$ Assume first that $\v_1=\v_2.$ Let  $u,v$ be integers satisfying \be\la{uv} um_1+vm_2=l.\ee Multiplying the
first equality in \eqref{100} by $u$ and adding
the second equality multiplied by $v$,
we see that $$l\alpha= \v_1l\beta + 2\pi k_1 u+2\pi k_2 v,$$  implying that
$T_{l}(a)=T_{l}(b)$.

Assume now that  $\v_1=-\v_2.$ Then, similarly, we conclude that
\be \la{qaz} l\alpha= \v_1\beta(um_1-vm_2) + 2\pi k_1 u+2\pi k_2 v.\ee
Furthermore, eliminating $\alpha$ from equalities \eqref{100} we obtain \be \la{vu} \v_1m_1m_2\beta=\pi k_2 m_1-\pi k_1 m_2.\ee Since
\be \la{pizz} T_n^{\prime}(\cos \phi)=n(\sin n\phi/\sin \phi),\ee equality \eqref{vu} implies that $T_{m_1m_2}^{\prime }(b)=0$, unless
$\beta=\pi k_3,$ $k_3\in \Z.$ In the last case, $b=1$ if $k_3$ is even, and $b=-1$ if $k_3$ is odd, implying that
$$T_{l}(b)=(-1)^{k_3l},$$ in view of \eqref{tyui}. On the other hand, if $\beta=\pi k_3$, then
\eqref{qaz} implies that
$$T_{l}(a)=(-1)^{k_3(u m_1-v m_2)}.$$ Since the sum and the difference of any two numbers have the same parity this implies that
$$T_{l}(a)=(-1)^{k_3(u m_1+v m_2)}=(-1)^{k_3l}=T_{l}(b).$$ Similarly, one can see that
$T_l(a)=T_l(b)$ unless $T_{m_1m_2}^{\prime }(a)=0.$

In order to prove b) observe that equalities \eqref{uuiioo} imply the equalities
\be \la{10} m_1 \alpha = \v_1 m_1 \beta + 2\pi k_1, \ \ \
m_2 \alpha = \v_2 m_2 \beta+ 2\pi k_2, \ \ \
m_3 \alpha = \v_3  m_3 \beta+ 2\pi k_3, \ee where $\v_1=\pm 1,$ $\v_2=\pm 1,$ $\v_3=\pm 1,$
and $k_1,k_2,k_3\in \Z.$
Clearly, among the numbers $\v_1,$ $\v_2,$ $\v_3$ at least two are equal
and we conclude as above that the equality $T_{l}(a)=T_{l}(b)$ holds for $l=\GCD(m_{i_1},m_{i_2})$, where $\v_{i_1}=\v_{i_2}.$
\qed

\bc \la{xyi}  Let $T_{m_1},$ $T_{m_2}$ be the Chebyshev polynomials  and $a$, $b$ be distinct complex numbers.
Assume that
$$T_{m_1}(a)=T_{m_1}(b), \ \ \ \ \ \ T_{m_2}(a)=T_{m_2}(b)$$ and $\GCD(m_1,m_2)=1$.
Then $$ T_{m_1m_2}^{\prime }(a)=T_{m_1m_2}^{\prime }(b)=0.$$
\ec
\pr Follows from Lemma \ref{skun}, a) taking into account that
the equality $S(a)=S(b)$ for some polynomial $S$ and $a\neq b$ obviously implies that $\deg S>1$. \qed

Recall that a number $\gamma\in \C$ is called algebraic if it is a root of an equation with rational coefficients. The set of all algebraic numbers is a subfield of $\C.$ A monic polynomial $p(x)\in \Q[x]$ of the minimal degree
such that $p(\gamma)=0$ is called a minimal polynomial of $\gamma.$ A minimal polynomial is irreducible over $\Q$. An algebraic number $\gamma$ is called an algebraic integer  if its minimal polynomial has integer coefficients. In fact, this condition may be replaced by a weaker condition
that $\gamma$ is a root of {\it  some}  monic polynomial with integer coefficients.
The set of all algebraic integers is closed under addition and multiplication.

\bl \la{poi} Assume that $a\in \C$ is a root of $T_n^{\prime}.$ Then $a\in \R$, and $2a$ is an algebraic integer.
\el
\pr Since equality \eqref{pizz} shows that $T^{\prime}_n$ has $n-1$ distinct real roots, all roots of
$T^{\prime}_n$ are real. The other statements follow
from the formulas $$T_n^{\prime}=nU_{n-1}$$ and
$$U_n=\sum_{k=0}^{[n/2]}(-1)^k \binom{n-k}{k}(2x)^{n-2k},$$
where $U_n$ denotes the Chebyshev polynomial of the second kind. \qed

\bc \la{red} In the notation of Theorem \ref{mpc} assume that $Q$ is a solution of \eqref{mom} of the second type, or
a solution of the third type,
which cannot be represented as a solution of the first type. Then $2V(a)$ and $2V(b)$ are algebraic integers.
\ec

\pr Without loss of generality we may assume that $V=x.$ If $Q$ is of the second type, then the statement follows from Corollary \ref{xyi} and Lemma \ref{poi}.

If $Q$ is of the third type, then applying
Lemma \ref{skun}, a) and Lemma \ref{poi}
to the equalities
$$T_{2m_1}(a)=T_{2m_1}(b), \ \ \ \ T_{2m_2}(a)=T_{2m_2}(b),$$ we conclude that $2V(a)$ and
$2V(b)$ are algebraic integers, unless  $T_2(a)=T_2(b)$. However,
the last equality yields the equality $a=-b,$ implying, as it was observed above, that
$Q$ can be represented as a solution of the first type. \qed

\subsection{Decompositions of polynomials with real coefficients}
In this subsection we collect necessary results concerning decomposition of polynomials with real coefficients into compositions of polynomials of lesser degree.

The following lemma is well known (see e. g. Corollary 2.2 in \cite{pak}).

\bl \la{wsx} Assume that $$P=A\circ B=\t A\circ \t B,$$ where $P, A,B,\t A,\t B\in \C[z]$ and
$\deg A=\deg \t A.$ Then there exists a polynomial $\mu\in \C[z]$ of degree one such that
$$\t A=A\circ \mu^{-1}, \ \ \t B=\mu\circ B.\ \ \ \Box$$
\el

\bc \la{reco0} Let $P=U\circ V,$ where $P\in \R[x],$ while $U,V\in \C[z].$
Assume that the leading coefficient of $V$ and its constant term are real numbers. Then
$U,V\in \R[x].$
\ec
\pr Since $P\in \R[x],$ we have:
\be \la{uip} P=U\circ V=\overline{U}\circ \overline{V},\ee where $\overline{U},$ $\overline{V}$ are polynomials obtained from $U,V$ by the complex conjugation of all coefficients. By Lemma \ref{wsx},  equality \eqref{uip} implies that
\be \la{iui} \overline{U}=U\circ \mu^{-1}, \ \ \overline{V}=\mu\circ V,\ee
where $\mu=\alpha z+\beta$ for some $\alpha, \beta\in \C.$ Since the leading coefficient of $V$ is real, the second equality in \eqref{iui}
implies that $\alpha=1$. Now the equality
$\overline{V}=V+\beta$ implies that $\beta=0,$ since the constant term of $V$ is real.
Therefore, $\overline{U}=U,$ $\overline{V}=V$ and
hence $U,V\in \R[x].$ \qed

\bc \la{reco} Assume that $$P=U\circ V,$$ where $P\in \R[x],$ while $U,V\in \C[z].$ Then there exists a polynomial $\mu\in \C[z]$ of degree one such that the polynomials
$$U_1=U\circ \mu^{-1}, \ \ V_1=\mu\circ V$$
are contained in $\R[x].$
\ec
\pr Let $\mu$ be any polynomial of degree one such that the leading coefficient and the constant term
of
the polynomial $V_1=\mu\circ V$ are real numbers.
Then $U_1$ and $V_1$ are contained in $\R[x]$ by Corollary \ref{reco0}. \qed

\bl \la{eche} Let $\mu_1,\mu_2$ be complex polynomials of degree one.
\vskip 0.1cm
\noindent a) Assume that the polynomial $\mu_1\circ z^n\circ \mu_2$, $n\geq 2,$ has real coefficients.
Then there exist $\t \mu\in \R[x]$ and $c\in \C$ such that
$\mu_2=c\t \mu.$
\vskip 0.1cm
\noindent b) Assume that the polynomial $\mu_1\circ T_n\circ \mu_2$, $n\geq 2,$ has real coefficients.
Then either $\mu_2\in \R[x]$ or
there exists $\t \mu\in \R[x]$ such that
$\mu_2=i\t \mu.$
\el
\pr
Let $\mu_1=\alpha_1z+\beta_1,$ $\mu_2=\alpha_2z+\beta_2,$ where $\alpha_1,\beta_1,\alpha_2,\beta_2\in \C.$ Then the coefficients of
$z^n$ and $z^{n-1}$ of the polynomial $\mu_1\circ z^n\circ \mu_2$ are $c_n=\alpha_1\alpha_2^n$ and
$c_{n-1}=\alpha_1\alpha_2^{n-1}\beta_2 n$, correspondingly. Since by assumption these numbers are real, we conclude that $$ \frac{c_{n-1}}{c_n}=\frac{n\beta_2}{\alpha_2}$$ is also a real number. Therefore,
$\beta_2/\alpha_2\in \R$, and hence
$\mu_2=\alpha_2\t \mu,$ where $\t \mu=x+(\beta_2/\alpha_2)\in \R[x].$

Similarly, since
$$T_n=2^{n-1}x^n-n2^{n-3}x^{n-2}+\dots$$ by \eqref{cheb}, the coefficients of
$z^n$, $z^{n-1}$, $z^{n-2}$ of the polynomial $\mu_1\circ T_n\circ \mu_2$ are: $$c_n=\alpha_12^{n-1}\alpha_2^n, \ \ \  c_{n-1}=\alpha_12^{n-1}n\alpha_2^{n-1}\beta_2, \ \ \ c_{n-2}=\alpha_12^{n-2}n(n-1)\alpha_2^{n-2}\beta_2^2-\alpha_12^{n-3}n\alpha_2^{n-2},
$$
correspondingly. As above, $c_n,c_{n-1}\in \R$ implies that $\beta_2/\alpha_2\in \R.$ Since
$$c_{n-2}=\frac{n(n-1)c_n}{2}\left(\frac{\beta_2}{\alpha_2}\right)^2-\frac{nc_n}{4}\left(\frac{1}{\alpha_2}\right)^2,$$
it follows now from $c_{n-2}\in \R$ that $\alpha_2^2\in \R$ implying the statement. \qed

\subsection{Solution of the polynomial moment problem over $\R$}
In this subsection we deduce from Theorem \ref{mpc} a description of polynomials
$P,$ $Q$ with {\it real} coefficients satisfying \eqref{mom} for  $a,b\in \R$.

The theorem below is a ``real'' analogue of Theorem \ref{mpc}.
Keeping the above notation we will call a solution $Q\in \R[x]$ of \eqref{mom} reducible if \eqref{c} holds
for some $\t P, \t Q, W\in \R[x].$ We also will call solutions 1), 2), 3)
described below solutions of the first, the second, and the third type.
Notice that the set of
solutions of the first type in the real case is ``smaller'' than the one in the complex case.

\bt \l{mpr} Let $P,$ $Q$ be non-constant real polynomials
and $a,b$ distinct real numbers such that equalities \eqref{mom} hold. Then, either $Q$ is a reducible solution of \eqref{mom}, or there exist
real polynomials $P_j,$ $Q_j,$ $V_j,$ $W_j,$ $1\leq j \leq r,$
such that
$$Q=\sum_{j=1}^rQ_j, \ \ \
P=P_j\circ W_j, \ \ \
Q_j=V_j\circ W_j, \ \ \  \ \ \ W_j(a)=W_j(b).
$$ Moreover,
one of the following conditions holds:
\vskip 0.25cm
\noindent 1) $r=2$ and
$$P=U\circ x^{2}R^2(x^2) \circ V,\ \ \ W_1=x^2 \circ V, \ \ \ W_2=xR(x^2)\circ V,$$
where $R,$ $U$, $V$  are real polynomials;
\vskip 0.25cm
\noindent 2)  $r=2$ and
$$P=U\circ T_{m_1m_2} \circ V,\ \ \ W_1= T_{m_1} \circ V, \ \ \
W_2= T_{m_2}\circ V,$$
where $U$, $V$ are  real polynomials, $m_1>1,$ $m_2>1$, $\GCD(m_1,m_2)=1;$
\vskip 0.25cm
\noindent 3) $r=3$ and
$$P=U\circ x^{2}R^2(x^{2})\circ  T_{m_1m_2}
\circ V,$$
$$W_1=T_{2m_1} \circ V, \ \ \ W_2=T_{2m_2}\circ V, \ \ \
W_3= (xR(x^2) \circ T_{m_1m_2}) \circ V,$$
where $R,$ $U$, $V$ are  real polynomials,
$m_1>1,$ $m_2>1$ are odd, and $\GCD(m_1,m_2)=1.$
\et
\pr Our strategy is to apply Theorem \ref{mpc} and to use the condition that $P,Q\in \R[x]$ and $a,b\in \R.$ Assume first that \eqref{c} holds for some $\t P, \t Q, W\in \C[z].$
Applying Corollary \ref{reco} to the equality $P=\t P\circ W$ we conclude  that without loss of generality we may assume that $\t P$ and $W$ are contained in $\R[x]$.  Now the equality $Q=\t Q\circ W$
implies by  Corollary \ref{reco0} that
$\t Q$ also is contained in $\R[x]$.

Assume that $Q$ is a solution of the first type. It follows from the equality $P=P_1\circ W_1$ by Corollary \ref{reco} that there exists a complex polynomial of degree one $\mu_1$ such that the polynomial $\mu_1\circ W_1$ has real coefficients. Further, applying
Corollary \ref{reco}  to
the equality $\mu_1\circ W_1=\mu_1\circ z^n\circ V$, we conclude that  there exists a complex polynomial of degree one $\mu_2$ such that the polynomials $\mu_1\circ z^n\circ \mu_2$ and $\mu_2^{-1}\circ V$ have real coefficients. By Lemma \ref{eche}, a) this implies that
there exist $\t \mu\in \R[x]$ and $c\in \C$ such that
$\mu_2=c\t \mu.$ Since $\mu_2^{-1}=\t\mu^{-1}\circ z/c,$ it follows now from $\mu_2^{-1}\circ V\in \R[x]$ that $V/c\in \R[x].$ Therefore, changing the polynomial $V$ to $V/c$, and modifying the polynomials
$P_1,$ $V_1,$ and $R$ in an obvious way, without loss of generality we may assume that $V\in \R[x].$

Clearly, $V\in \R[x]$ implies that $W_1=z^n\circ V\in \R[x]$. It follows now from  $P=P_1\circ W_1$
by Corollary \ref{reco0} that $P_1\in \R[x]$. Furthermore,
it follows from $W_1(a)=W_1(b)$ and $a,b\in \R$ that $n=2k$, and $V(a)=-V(b).$
Since $\GCD(s,n)=1,$ the polynomial $z^sR(z^n)$ has the form $z\t R(z^2)$ for some $\t R\in \C[z]$.
Thus, changing $P_1$ tp $P_1\circ z^{n/2}$ and $z^sR(z^n)$ to $z\t R(z^2)$,
without loss of generality we may assume that $W_1=z^2\circ V$ and
$W_2=zR(z^2)\circ V$. Applying
Corollary \ref{reco0} to the equality $P=(P_2\circ zR(z^2))\circ V$ we see that $P_2\circ zR(z^2)\in \R[x]$. Therefore, taking into account that the constant term of $zR(z^2)$ is zero,
Corollary \ref{reco0} implies that for $c\in \C$ such that the leading coefficient of
$czR(z^2)$ is real the polynomials $P_2\circ z/c$ and  $czR(z^2)$ are contained in $\R[x]$.
Thus, modifying the polynomials $P_2$ and $R$ we can assume that they are
contained in $\R[x]$. Now Corollary \ref{reco0} applied to the equality $P=U\circ (z^2R^2(z^2)\circ V)$ implies that $U\in \R[x]$.

Finally, the equality
$$Q=V_1\circ W_1+V_2\circ W_2=\overline{V}_1\circ W_1+\overline{V}_2\circ W_2$$ implies that
$$Q=\frac{V_1+\overline{V}_1}{2}\circ W_1+\frac{V_2+\overline{V}_2}{2}\circ W_2.$$
Therefore, changing if necessary $V_1$ to $(V_1+\overline{V}_1)/2$ and $V_2$ to $(V_2+\overline{V}_2)/2,$ without loss of generality we may assume that $V_1,$ $V_2\in \R[x].$

Assume now that $Q$ is a solution of the third type. We may assume that $V(a)\neq - V(b)$, for otherwise, as it was observed after Theorem \ref{mpc}, this solution also belongs to the first type considered earlier.
As above, we conclude that there
exist complex polynomials of degree one $\mu_1$ and $\mu_2$ such that the polynomials $\mu_1\circ T_{2m_1}\circ \mu_2$ and $\mu_2^{-1}\circ V$ have real coefficients. By Lemma \ref{eche}, b), this implies that
either $\mu_2\in \R[x]$ or
there exists $\t \mu\in \R[x]$ such that
$\mu_2=i\t \mu.$ Since, $\mu_2^{-1}\circ V\in \R[x],$ in the first case $V\in \R[x],$ while in the second one, \be \l{asdf} V=i\t V, \ \ \ \t V\in \R[x].\ee Let us show that equality \eqref{asdf} is impossible.
Indeed, applying Lemma \ref{skun}, a) to the equalities $W_1(a)=W_1(b),$ $W_2(a)=W_2(b)$ and arguing as in Lemma \ref{red}, we conclude that the numbers $V(a)$ and $V(b)$ are roots of the polynomial $T_{4m_1m_2}^{\prime}$, for otherwise $V(a)= - V(b)$. Since
$T^{\prime}_n$ has only real zeroes, we conclude that $V(a),V(b) \in \R$, and hence \eqref{asdf} is impossible in view of $a,b\in \R.$
Thus, $V\in \R[x]$.

Applying now Corollary \ref{reco0} to the equality $P=(P_3\circ zR(z^2))\circ (T_{m_1m_2}\circ V)$ we conclude that $P_3\circ zR(z^2)\in \R[x]$. Furthermore, arguing as above, we conclude that without loss of generality we may assume that
$P_3,R\in \R[x]$ as well as $P_1,P_2,U\in \R[x]$ and $V_1,V_2,V_3\in \R[x].$

The
proof of the theorem in the case where $Q$ is a solution of the second type is obtained similarly with obvious simplifications. \qed

\section{Proof of Theorem \ref{t2}}
\subsection{Plan of the proof}
In the rest of the paper we always will assume that all considered polynomials  have real coefficients.

Let us describe a general plan of the proof of Theorem \ref{t2}. First, observe that we may assume that
\be \la{lu} \R(P,Q)=\R(x).\ee Indeed, otherwise by the L\"uroth theorem, $\R(P,Q)=\R(W)$ for some $W\in \R(x),$ $\deg W\geq 2,$ implying that
$$P=\t P \circ W, \ \ \ Q=\t Q\circ W$$ for some $\t P, \t Q \in \R(x)$ such that $\R(\t P,\t Q)=\R(x).$ Moreover, since $P,Q\in \R[x]$, it is easy to see that
we may assume that $\t P, \t Q, W\in\R[x].$ Therefore, since equalities  \eqref{mix} imply the equalities
\be
\int_{W(a)}^{W(b)} \t P^id \t Q=0, \ \ \  \int_{W(a)}^{W(b)} \t Q^jd \t P=0, \ \ \ i,j \geq 0,
\ee
it is enough to prove that if  \eqref{lu} holds, then $P,Q$ cannot satisfy \eqref{mix} for  $a\neq b.$

Applying Theorem \ref{mpr} to the first and to the second system of equations in \eqref{mix} separately, we arrive to nine different
``cases'' depending on types of solutions appearing in Theorem \ref{mpr}. For example, ``the case (2,1)'' means that $Q$ is a solution of the second type
of the polynomial moment problem \eqref{mom}, while $P$ is a solution of the first type of the polynomial moment problem
$$\int_a^b Q^jd P=0, \ \ \ j \geq 0.$$
In more details, this means that, from one hand,
$$Q=V_1\circ  W_1+ V_2\circ  W_2,\ \ \ P=U\circ T_{nm} \circ  V,$$ where
\be \la{uio}  W_1=T_n \circ  V, \ \ \ W_2=T_m\circ  V,\ee and
$$W_1(a)=W_1(b), \ \ \ W_2(a)=W_2(b),$$ while, from the other hand,
$$P=\t V_1\circ \t W_1+\t V_2\circ \t W_2 ,\ \ \ Q=\t U\circ  x^{2} R^2(x^2)\circ\t V,$$ where

$$\t W_1= x^2 \circ \t V, \ \ \
\t W_2=xR(x^2)\circ \t V,$$  and
$$\t W_1(a)=\t W_1(b), \ \ \ \t W_2(a)=\t W_2(b).$$
In view of assumption \eqref{lu}, the polynomial $V$ (as well as the polynomial $\t V$) is of degree one
for otherwise
$$\R(P,Q)\subseteq\R(V)\subsetneq \R(x).$$ Furthermore, it is clear that without loss of generality we may assume that one of the polynomials $V$ and $\t V$ equals $x.$
Our strategy will be to show that such systems of equations always imply that equalities \eqref{c} hold, in contradiction with \eqref{lu} (recall that
the condition $W(a)=W(b)$  implies that $\deg W>1$).

Since we may exchange $P$ and $Q$, it is necessary to consider only the cases (1,1), (2,1), (2,2), (3,1), (3,2), and (3,3).
Finally,
we may impose some additional restrictions related to the fact that  a solution  of the (usual) polynomial moment problem may belong to different types. For example, assuming that the theorem is already proved in the case (1,1), considering the case (2,1) we may assume that $n>2,$ $m>2$ in \eqref{uio}, since otherwise the solution
$P,Q$  also  belongs to the case (1,1).

For a polynomial $$P=a_nx^n+a_{n-1}x^{n-1}+ ... +a_1x+a_0, \ \ \ a_i\in \R, \ \  \ 0\leq i \leq n,$$ of degree $n$, set $$C_i(P)=a_{n-i}, \ \ \ 0\leq i \leq n.$$

The following simple lemma permits to control initial terms in a composition
of two polynomials and is widely used in the following.

\bl \la{l1} Let $T$ be a polynomial of degree $d$.
Then for any polynomial $S$ of degree $r$ with the leading coefficient $c$ the equalities
\be \la{wer} C_i(S\circ T)=C_i(cx^r\circ T), \ \ \ 0\leq i \leq d-1,\ee hold. In particular,
for any two polynomials $S_1,$ $S_2$ of equal degree  with equal leading coefficients
the equalities
$$ C_i(S_1\circ T)=C_i(S_2\circ T), \ \ \ 0\leq i \leq d-1,$$ hold.
\el
\pr Indeed, $\deg (S-cx^r)\circ T=dr-d$. Therefore, \eqref{wer} holds. \qed

\bc \la{c1} Let $T(z)$ be a polynomial of degree $d\geq 2$  such that  $C_1(T)=0$ holds,
$U$ be an arbitrary polynomial, and $\alpha, \beta\in \R,$ $\alpha\neq 0.$ Then  equality $C_1(U\circ T\circ (\alpha x+\beta))=0$ holds if and only if $\beta=0.$
\ec
\pr Indeed,   if $\deg U=r$, $C_0(U)=c$, then
$C_1(U\circ T)=C_1(cz^r\circ T)$ by Lemma \ref{l1}. On the other hand, $C_1(cz^r\circ T)=0$, since $C_1(T)=0$. Therefore, $C_1(U\circ T)=0$, and it is easy to see that for any polynomial $F$ such that  $C_1(F)=0,$ the equality
$C_1(F\circ (\alpha x+\beta))= 0$  holds if and only if $\beta=0.$
\qed

\subsection{Proof of Theorem \ref{t2} in the cases (1,1)}

\bl \la{ll2} Let $W_1$, $W_2$ be polynomials of degree two such that
$W_1(a)=W_1(b),$ $W_2(a)=W_2(b)$ for distinct $a,b\in \R$. Then $W_2=\lambda_1W_1+\lambda_2$ for some $\lambda_1,$ $\lambda_2 \in \R.$
\el
\pr
Let
$$P=\alpha_1x^2+\beta_1x+\gamma_1, \ \ \ \ Q=\alpha_2x^2+\beta_2x+\gamma_2,$$
where $\alpha_1,\beta_1,\gamma_1,\alpha_2,\beta_2,\gamma_2\in \R.$
Then conditions of the lemma yield the equalities
$$\alpha_1(a+b)+\beta_1=0, \ \ \ \alpha_2(a+b)+\beta_2=0,$$ implying the statement.
\qed
In order to proof the theorem in the case (1,1) it is enough to observe that in this case there exist $U,R,\t U, \t R\in \R[x]$ such that
$$P=(U\circ zR(z^2))\circ W_1, \ \ \ Q=(\t U\circ z\t R(z^2))\circ \t W_1$$
where
$$W_1=z^2\circ V, \ \ \ \ \t W_1= z^2\circ \t V,$$
are polynomials of degree two such that
$W_1(a)=W_1(b),$ $\t W_1(a)=\t W_2(b)$. By Lemma \ref{ll2}, we have
$\t W_2=\lambda_1W_1+\lambda_2$ for some $\lambda_1,$ $\lambda_2 \in \R$, and hence
\eqref{c} holds for $W=W_1.$

\subsection{Proof of Theorem \ref{t2} in the case (2,1)}
If $P,Q$ is a solution of \eqref{mix} corresponding to the case (2,1), then without loss of generality we may assume that
there exist polynomials $V_1,V_2, U, \t V_1,\t V_2, \t U,  R$ and $\alpha,\beta\in \R,$ $\alpha\neq 0,$ such that
\be \la{urab} Q=V_1\circ T_{m_1}+V_2\circ T_{m_2}
=\t U\circ x^2{R}^2(x^2)\circ (\alpha x+\beta),\ee and
$$ P=\t V_1\circ x^2 \circ (\alpha x+\beta)+\t V_2\circ xR(x^2) \circ (\alpha x+\beta)
=U\circ T_{m_1m_2} ,$$
where  $\GCD(m_1,m_2)=1.$ In addition, for
the polynomials $$ W_1=T_{m_1}, \ \ W_2=T_{m_2}, \ \ \t W_1=z^2\circ (\alpha x+\beta),\ \  \t W_2=xR(x^2)\circ (\alpha x+\beta)$$
the equalities \be \la{ko2} W_1(a)=W_1(b), \ \ \ W_2(a)=W_2(b), \ \ \ \t W_1(a)=\t W_1(b), \ \ \ \t W_2(a)=\t W_2(b)\ee hold.
If at least one of the numbers $m_1,m_2$ equals 2, then $P,Q$ belongs to the type (1,1) considered above.
So, we may assume that  $m_1\geq 3,$ $m_2\geq 3$. Notice that the second representation for $Q$ in \eqref{urab} implies that $n=\deg Q$ is even and the inequality $n\geq 6$ holds.

Although the above conditions seem to be very strong, it is difficult to use them in their full generality since they contain many unknown parameters. Thus, actually we mostly will use only the fact that the right part of  \eqref{urab} is a polynomial in $x^2\circ (\alpha x+\beta)$ together with first three equalities in \eqref{ko2}.

First of all observe that any polynomial of the form $Q=V_1\circ T_{m_1}+V_2\circ T_{m_2}$ can be represented in the form  \be \la{xy} Q=d_nT_n+d_{n-1}T_{n-1}+\dots +d_{1}T_1+d_0,\ \ \ \ \ d_i\in \R,\ee
where $d_i=0$, unless $i$ is divisible either by $m_1$ or by $m_2$.
Indeed, it is clear that $T_0,T_1, \dots , T_{r}$ is a basis of a subspace of $\R[x]$ consisting of all polynomials of degree $\leq r$.
Therefore, a polynomial $P$ can be represented in the form
$$P=V\circ T_m=(a_rx^r+a_{r-1}x^{r-1}+...+a_1x+a_0)\circ T_{m}$$ if and only if
$$P=(b_rT_r+b_{r-1}T_{r-1}+ ... +b_1T_1+b_0)\circ T_m=b_rT_{rm}+b_{r-1}T_{(r-1)m}+ ... +b_1T_m+b_0.$$ Consequently,
$Q=V_1\circ T_{m_1}+V_2\circ T_{m_2}$ can be represented in the required form.

Define $C(n,m_1,m_2)$ as the set of all polynomials \eqref{xy} such that $d_i=0$, unless $i$ is divisible either by $m_1$ or by $m_2$, and $d_n\neq 0.$
To be definite, we always will assume that $n$ is divisible by $m_1.$
Similarly to the notation $C_i(Q)$ introduced above, for a polynomial $Q\in C(n,m_1,m_2)$ set
$$Ch_i(Q)=d_{n-i}, \ \ \ 0\leq i \leq n.$$

\bl \la{dfg} Let $Q\in C(n,m_1,m_2)$, where $m_1$ and $m_2$ are
coprime, $m_1\geq 3,$ $m_2\geq 3$.

\vskip 0.2cm
\noindent 1) If $Ch_1(Q)\neq 0$, then $Ch_2(Q)=0.$
\vskip 0.2cm
\noindent 2) If $Ch_1(Q)\neq 0,$ $Ch_3(Q)\neq 0,$ then $m_1=3$, and $Ch_4(Q)= 0,$
\vskip 0.2cm
\noindent 3)  If $Ch_1(Q)\neq 0,$ $Ch_3(Q)\neq 0,$ $Ch_5(Q)\neq 0,$ then $m_1=3,$ $m_2=4.$
\el
\pr
If $Ch_1(Q)\neq 0$, then  $m_2\mid n-1.$
Further, clearly
\be \la{inv} \vert m_ik_1-m_ik_2\vert\geq m_i\geq 3, \ \ \ i=1,2,\ \ \ k_1,k_2\in \N,\ee   unless $k_1=k_2$,
implying that
$n-2$ may not be divisible neither by $m_1$, since $m_1\mid n$,
nor  by $m_2$, since $m_2\mid n-1$.
Thus, $Ch_2(Q)=0.$

Assume that additionally $Ch_3(Q)\neq 0$. Since $m_2\mid n-1$, the number $n-3$ may not be divisible by $m_2$ in view of
\eqref{inv}. Therefore, $n-3$ is divisible by $m_1$ implying that $m_1=3$. Furthermore, $Ch_4(Q)=0$ for otherwise \eqref{inv} implies that $m_2=3$ in contradiction with $\GCD(m_1,m_2)=1.$

Finally, if additionally $Ch_5(Q)\neq 0$, it follows from $m_1=3$ by \eqref{inv}
that $m_2\vert n-5$ implying that $m_2=4.$ \qed

\bc \la{ddfgg} Let $Q\in C(n,m_1,m_2)$, where $m_1$ and $m_2$ are
coprime, $m_1\geq 3,$ $m_2\geq 3$, and $n$ is even. If $Ch_1(Q)\neq 0$, then
either $Ch_2(Q), Ch_3(Q)$ vansih, or $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish.

\ec
\pr Since $m_2\vert n-1$, and $n$ is even, $m_2$ is odd. Therefore, $m_2\neq 4$, and the statement follows from
Lemma \ref{dfg}. \qed

\bl \la{l2} Let $Q=F\circ x^2\circ (x-\delta)$, $\delta \in \R,$ where $n=\deg Q\geq 6.$
\vskip 0.2cm
\noindent 1) If the coefficients $Ch_2(Q)$, $Ch_3(Q)$ vanish, then either $\delta=0,$ or \be \la{azx} (2\delta)^2=\frac{3}{(n-1)(n-2)}. \ee
\vskip 0.2cm
\noindent 2) If the coefficients $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish, then either $\delta=0,$ or $2\delta$ satisfies the equation
\be \la{xza} \frac{2}{15}(n-1)(n-2)(n-3)(n-4)t^4-(n-2)(n-3)t^2+1=0.\ee
\el
\pr If $Ch_2(Q)$, $Ch_3(Q)$ vanish, then $Q=c_0T_n+c_1T_{n-1}+ R_{1},$ where $\deg R_1\leq n-4$.
Set $$T_s^*(x)=2T_s\left(\frac{x}{2}\right), \ \ \ s\geq 1.$$ Clearly, the equality
$$ c_0T_n+c_1T_{n-1}+ R_{1}=F\circ x^2\circ (x-\delta)$$ implies the equality
\be \la{0100} c_0T_n^*+c_1T_{n-1}^*+ \widetilde R_{1}=\t F\circ x^2\circ (x-\gamma),\ee
where $\t R_1=2R_1(x/2),$ $\t F=2F(x/4),$ and
$\gamma=2\delta.$
Furthermore, without loss of generality we may assume that $c_0=1.$

Since the right part of \eqref{0100} is a polynomial in $(x-\gamma)^2$, it follows from  \eqref{0100} taking into account the inequality $\deg R_1\leq n-4$
that the derivatives of $T_n^*+c_1T_{n-1}^*$ of orders $n-1$ and $n-3$ at the point $\gamma$ vanish, that is
$$T_n^{*(n-1)}(\gamma)+c_1T_{n-1}^{*(n-1)}(\gamma)=0,$$
\vskip -0.2cm
$$T_n^{*(n-3)}(\gamma)+c_1T_{n-1}^{*(n-3)}(\gamma)=0.$$
Since
\be \la{f} T_s^*=x^s-sx^{s-2}+\frac{s(s-3)}{2}x^{s-4}-\frac{s(s-4)(s-5)}{6}x^{s-6}+ ... \ee by \eqref{cheb}, this implies that
$$n!\gamma+c_1(n-1)!=0,$$
$$\frac{n!}{3!}\gamma^3-n(n-2)!\gamma+c_1\left(\frac{(n-1)!}{2!}\gamma^2-(n-1)(n-3)!\right)=0.$$
The first of these equalities implies
that $c_1=-n\gamma$. Substituting this value of $c_1$ in the second equality we obtain
$$\frac{n!}{3!}\gamma^3-n(n-2)!\gamma-n\gamma\left(\frac{(n-1)!}{2!}\gamma^2-(n-1)(n-3)!\right)=-\frac{n!}{3}\gamma^3+n(n-3)!\gamma=0$$ implying that $2\delta= \gamma$ satisfies \eqref{azx}, unless $\delta=0.$

Similarly, if $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish, we arrive to the equality
\be \la{1} T_n^*+c_1T_{n-1}^*+c_3T_{n-3}^*+ \widetilde R_{1}=\t R\circ x^2\circ (x-\gamma),\ee
where $\deg \t R_1\leq n-6$ and $\gamma=2\delta,$ implying
that the derivatives of $T_n^*+c_1T_{n-1}^*+c_3T_{n-3}^*$ of orders $n-1$ and $n-3,$ and $n-5$ at the point $\gamma$ vanish. Thus,
$$T_n^{*(n-1)}(\gamma)+c_1T_{n-1}^{*(n-1)}(\gamma)=0,$$
\vskip -0.2cm
$$T_n^{*(n-3)}(\gamma)+c_1T_{n-1}^{*(n-3)}(\gamma)+c_3T_{n-3}^{*(n-3)}(\gamma)=0,$$

$$T_n^{*(n-5)}(\gamma)+c_1T_{n-1}^{*(n-5)}(\gamma)+c_3T_{n-3}^{*(n-5)}(\gamma)=0,$$
or equivalently
$$n!\gamma+c_1(n-1)!=0,$$
$$\frac{n!}{3!}\gamma^3-n(n-2)!\gamma+c_1\left(\frac{(n-1)!}{2!}\gamma^2-(n-1)(n-3)!\right)+c_3(n-3)!=0,$$ and
$$\frac{n!}{5!}\gamma^5-\frac{n(n-2)!}{3!}\gamma^3+\frac{n(n-3)(n-4)!}{2}\gamma+$$ $$+c_1\left(\frac{(n-1)!}{4!}\gamma^4 -\frac{(n-1)(n-3)!}{2!}\gamma^2+\frac{(n-1)(n-4)(n-5)!}{2}\right)+$$
$$+c_3\left(\frac{(n-3)!}{2!}\gamma^2-(n-3)(n-5)!\right)=0.$$
As above, it follows from the first of these equalities by \eqref{f} that $c_1=-n\gamma$ and substituting this value of $c_1$ in the second equality we obtain
$$\frac{n!}{3!}\gamma^3-n(n-2)!\gamma-n\gamma\left(\frac{(n-1)!}{2!}\gamma^2-(n-1)(n-3)!\right)+c_3(n-3)!=0$$ implying that
$$c_3=\frac{n(n-1)(n-2)}{3}\gamma^3-n\gamma.$$
Now the third equality gives us
$$\frac{n!}{5!}\gamma^5-\frac{n(n-2)!}{3!}\gamma^3+\frac{n(n-3)(n-4)!}{2}\gamma-n\gamma\left(\frac{(n-1)!}{4!}\gamma^4 -\frac{(n-1)(n-3)!}{2!}\gamma^2+\frac{(n-1)(n-4)(n-5)!}{2}\right)+$$
$$+\left(\frac{n(n-1)(n-2)}{3}\gamma^3-n\gamma\right)\left(\frac{(n-3)!}{2!}\gamma^2-(n-3)(n-5)!\right)=0.$$
The coefficient of $\gamma^5$ in this expression is
$$\frac{n!}{3!}\left(\frac{1}{20}-\frac{1}{4}+1\right)=\frac{2n!}{15}\ .$$
The coefficient of $\gamma^3$  is
$$-\frac{n(n-2)!}{3!}+\frac{n(n-1)(n-3)!}{2!}-\frac{n(n-1)(n-2)(n-3)(n-5)!}{3}-\frac{n(n-3)!}{2!}=$$
$$= -\frac{n(n-3)!}{2!}\left(\frac{n-2}{3}-(n-1)\right)-n(n-3)(n-5)!\left(\frac{(n-1)(n-2)}{3}+\frac{(n-4)}{2}\right)=$$
$$=\frac{(2n-1)n(n-3)!}{6}-n(n-3)(n-5)!\left(\frac{2n^2-3n-8}{6}\right)=$$ $$=\frac{n(n-3)(n-5)!}{6}\left((2n-1)(n-4)-(2n^2-3n-8)\right)=-n(n-2)(n-3)(n-5)!$$

Finally, the coefficient of $\gamma$ is
$$\frac{n(n-3)(n-4)!}{2}-\frac{n(n-1)(n-4)(n-5)!}{2}+n(n-3)(n-5)!=
$$
$$=n(n-5)!\left(\frac{(n-3)(n-4)}{2}-\frac{(n-1)(n-4)}{2}+(n-3)\right)=n(n-5)!\, .
$$
Collecting terms and canceling by $n(n-5)!$ we see that $2\delta=\gamma$ satisfies
\eqref{xza}, unless $\delta=0.$ \qed

\bc \la{lal} Let $Q=R\circ x^2\circ (x-\delta)$, $\delta \in \R.$
\vskip 0.2cm
\noindent 1) If the coefficients $Ch_2(Q)$, $Ch_3(Q)$ vanish and $n=\deg Q\geq 6,$ then the number $4 \delta$
is not an  algebraic integer, unless $\delta=0.$

\vskip 0.2cm
\noindent 2) If the coefficients $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish and $n=\deg Q\geq 9,$
then the number $4 \delta$
is not an  algebraic integer, unless $\delta=0.$
\ec
\pr
Set $\gamma=4\delta.$ If $Ch_2(Q)$, $Ch_3(Q)$ vanish and $\delta\neq0,$ then $\gamma$ is a root of the equation
$$t^2-\frac{12}{(n-1)(n-2)}=0.$$
Since for $n\geq 6$ the number $\frac{12}{(n-1)(n-2)}$ is not an integer, this implies that  $\gamma$ cannot be an  algebraic integer
of degree two. Moreover, $\gamma$ cannot be an  algebraic integer
of degree one for otherwise $\gamma$ is an integer implying that $\gamma^2=\frac{12}{(n-1)(n-2)}$ is also an integer.

If $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish and $\delta\neq0,$ then $\gamma$ is a root of the equation
\be \label{ur} (n-1)(n-2)(n-3)(n-4)t^4-30(n-2)(n-3)t^2+120=0.\ee
If this equation is irreducible over $\Q$, then $\gamma$ cannot be an algebraic integer since
the number $(n-1)(n-2)(n-3)(n-4)$ obviously does not divide 120 for $n\geq 9.$

Assume now that $\gamma$ is an algebraic integer satisfying an irreducible equation $t^2+c_1t+c_2=0,$ $c_1,c_2\in \Z.$ Then
by the Gauss lemma the equality
$$(n-1)(n-2)(n-3)(n-4)t^4-30(n-2)(n-3)t^2+120=(t^2+c_1t+c_2)(d_0t^2+d_1t+d_2)$$ holds for some $d_0,d_1,d_2\in \Z$.
Since the coefficients of  $t^3$ and $t$ in the left part vanish we have:
$$d_1+c_1d_0=0, \ \ \ \ c_1d_2+c_2d_1=0,$$ implying that \be \la{bege1} d_1=-c_1d_0, \ \ \ d_2=c_2d_0,\ee unless
\be \la{bege2} c_1=0,\ \ \  d_1=0.\ee

If \eqref{bege1} holds,  then
$120=c_2d_2=c_2^2d_0$ in contradiction with
$d_0=(n-1)(n-2)(n-3)(n-4)$ and $n\geq 9.$
Similarly, if \eqref{bege2} holds,
then $\gamma^2=-c_2$ is an integer, and
it follows from  \eqref{ur} that the number
$(n-2)(n-3)\gamma^2$ divides 120 implying easily a contradiction with the condition $n\geq 9.$

Finally, observe that if $\gamma$ is a rational root of $\eqref{ur}$, then $-\gamma$ also is a root of \eqref{ur} and $t^2-\gamma^2$ is a divisor of
\eqref{ur}. Therefore, any irreducible over $\Q$ factor of \eqref{ur} has the degree one, two, or four. Thus, in order to finish the proof it is enough
to observe
that $\gamma$ is not an integer, since otherwise $\gamma^2$ is also an integer implying as above that
$(n-2)(n-3)\gamma^2$ divides 120. \qed

\noindent{\it Proof of Theorem \ref{t2} in the case (2,1).}
Observe first that if $\beta=0$ in \eqref{urab}, then the theorem is true. Indeed, in this case the condition $\t W_1(a)=\t W_1(b)$ is equivalent
to the condition $T_2(a)=T_2(b).$
Therefore, applying Lemma \ref{skun}, b) to the equalities
$$T_{m_1}(a)=T_{m_1}(b), \ \ \ T_{m_2}(a)=T_{m_2}(b)\ \ \ T_2(a)=T_2(b),$$
we conclude that at least one of the numbers $m_1,$ $m_2$ is even and hence $Q=\t U\circ xR^2(x)\circ (\alpha x)^2 $ and
$P=U\circ T_{m_1m_2}=U\circ T_{m_1m_2/2}\circ T_2$ satisfy \eqref{c} for $W=x^2.$

Further, observe that \eqref{urab} implies that
\be\la{svina} Q=\t U\circ x{R}^2(x)\circ x^2\circ (\alpha x+\beta)=F\circ x^2 \circ \left(x+\frac{\beta}{\alpha}\right),\ee
where $F=\t U\circ x{R}^2(x)\circ \alpha^2 x,$ while the condition $\t W_1(a)=\t W_1(b)$ yields that
\be \la{a+b} a+b=-\frac{2\beta}{\alpha}.\ee

If $Ch_1(Q)=0$, then \eqref{xy} implies that also $C_1(Q)=0$, since $C_1(T_n)=0$ by \eqref{cheb}.
In its turn $C_1(Q)=0$ implies that $\beta=0$ by Corollary \ref{c1} applied to \eqref{svina}. So,
assume that $Ch_1(Q)\neq 0$. By Corollary \ref{ddfgg}, this implies that
either $Ch_2(Q), Ch_3(Q)$ vanish, or $Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish.

If $Ch_2(Q), Ch_3(Q)$ vanish, then
Corollary \ref{lal}, 1)  applied to \eqref{svina}  implies that  the number $-4 \beta/\alpha$
if  not an  algebraic integer, unless $\beta=0$. On the other hand, \eqref{a+b} implies that
$-4 \beta/\alpha$ is an  algebraic integer, since $2a$ and $2b$ are algebraic integers by Corollary \ref{red}.
Thus, we conclude again that $\beta=0.$

Similarly, assuming that $Ch_3(Q)\neq 0$, while
$Ch_2(Q)$, $Ch_4(Q),$ $Ch_5(Q)$ vanish,
we may apply Corollary \ref{lal}, 2),
whenever $n\geq 9.$ Therefore,
since $Ch_3(Q)\neq 0$ implies that $3\vert n$ in view of the equality $m_1=3$,
the only case
which remains uncovered is the one where $n=6$. In this case the inequality $Ch_1(Q)\neq 0$ implies that $m_2=5.$
Notice that it follows from \eqref{ur} that for $n=6$ the number  $4 \beta/\alpha$ satisfies the equation $t^4-3t^2+1=0$ whose roots
are algebraic integers. Moreover, the system
$$T_5(a)=T_5(b), \ \ T_3(a)=T_3(b), \ \  (2a+2b)^4-3(2a+2b)^2+1=0$$
has non-zero solutions. Thus, for $n=6$ the previous reasoning fails.

In order to prove the theorem in this case remind that in Lemma \ref{l2} we used a condition which is weaker than the one in \eqref{urab}. Therefore, in order to finish the proof
it is enough to show that
the equality
\be \la{tgv} T_6+c_1T_5+c_3T_3+c_6=c(z(z^2-d))^2\circ (z-\beta),\ee
where $c_1,c_3,c_6,c,d,\beta\in \R,$ is possible only if $c_1=0$.
This statement may be verified by
a direct calculation. Namely, the comparison of leading coefficients of both parts of \eqref{tgv} implies that $c=32,$ while the comparison of other coefficients gives
$$16 c_1+192 b=0, \ \ \  -480 b^2+64 d-48=0,$$ $$640 b^3-256 b d-20 c_1+4c_3=0, \ \ \ -480 b^4+384 b^2 d-32 d^2+18=0,$$ $$192 b^5-256 b^3 d+64 b d^2+5 c_1-3 c_3,  \ \ \  c_6=-32 b^6+64 b^4 d-32 b^2 d^2-1.
$$
We leave the reader to verify (for example, with the help of Maple) that the only solution of the above system is
$$
c_1 = 0,\ \  c_3 = 0,\  \ c_6 = 1, \  \ b = 0, \ \ d = 3/4,$$
(for these values of parameters equality
\eqref{tgv} simply reduces to the equality $T_6=T_2\circ T_3$).

\subsection{Proof of Theorem \ref{t2} in the case (2,2)}
First, observe that Theorem \ref{t2} in the case (2,2) follows from the following statement.

\bp \la{vot} Let $V_1,V_2, U, \t V_1, \t V_2, \t U\in \R[x]$ and $\alpha,\beta\in \R$, $\beta\neq 0,$ satisfy the equalities
\be \la{uraba} Q=V_1\circ T_{m_1}+V_2\circ T_{m_2}=\t U\circ T_{\t m_1\t m_2}\circ (\alpha x+\beta),\ee and
\be \la{uraba2} P=\t V_1\circ T_{\t m_1}\circ (\alpha x+\beta)+\t V_2\circ T_{\t m_2}\circ (\alpha x+\beta)=U\circ T_{m_1m_2},\ee
where \be \la{kvak} \GCD(m_1,m_2)=1, \ \ \ \GCD(\t m_1,\t m_2)=1,\ee and $m_1\geq 3, m_2\geq 3, \t m_1\geq 3,\t m_2\geq 3.$
Then $\alpha=\pm 1,$ $\beta =0.$

\ep

Indeed, in the case (2,2) conditions \eqref{uraba}, \eqref{uraba2}, \eqref{kvak} are satisfied for some $m_1\geq 2,$ $m_2\geq 2,$ $ \t m_1\geq 2,$ $\t m_2\geq 2.$
Additionally, the equalities
\be \la{asdd-} T_{m_1}(a)=T_{m_1}(b), \ \ \ T_{m_2}(a)=T_{m_2}(b),\ee
\be \la{asdd} T_{\t m_1}(\alpha a+\beta)=T_{\t m_1}(\alpha b+\beta), \ \ \
T_{\t m_2}(\alpha a+\beta)=T_{\t m_2}(\alpha b+\beta)\ee
hold.
If at least one of the numbers $m_1,m_2$ equals 2, then $P,Q$ belongs to the type (1,2) considered above. So, we may assume that  $m_1\geq 3,$ $m_2\geq 3$. Similarly, we may assume that
$\t m_1\geq 3,\t m_2\geq 3,$ since otherwise $P,Q$ belongs to the type (2,1).
Since under these conditions Proposition \ref{vot} implies that
$\alpha=\pm 1,$ $\beta =0,$ it follows from the equalities
$$T_{m_1}(a)=T_{m_1}(b), \ \ \ T_{m_2}(a)=T_{m_2}(b), \ \ \ T_{{\t m}_1}(\pm a)=T_{{\t m}_1}(\pm b)$$
by  Lemma \ref{skun}, b), taking into account \eqref{kots}, that
$T_s(a)=T_s(b)$, where either $s=\GCD(m_1, {\t m}_1)$, or $s=\GCD(m_2, {\t m}_2).$
Since
$$Q=\t U\circ T_{\t m_1\t m_2}\circ (\pm z)=\t U\circ (\pm  T_{\t m_1\t m_2})=\t U\circ (\pm  T_{\t m_1\t m_2/s})\circ T_s$$
and
$$P=U\circ  T_{m_1m_2}=U\circ  T_{m_1m_2/s}\circ T_s,$$
we conclude that \eqref{c} holds for $W=T_s.$

\vskip 0.2cm

The next lemmas are similar to Lemma \ref{dfg} and Lemma \ref{l2} and are used for imposing restrictions on possible values of $\alpha$ and $\beta$ in \eqref{uraba}, \eqref{uraba2}, and eventually to show that $\alpha=\pm 1,$ $\beta =0.$

\bl \la{l4}  Let $Q=F\circ T_s\circ (\alpha x+\beta)$, where $s\geq 5,$ $\alpha, \beta \in \R$, $\alpha\neq 0.$
\vskip 0.2cm
\noindent 1) If the coefficients $Ch_2(Q)$, $Ch_3(Q)$ vanish, then either $\alpha=\pm 1,$ $\beta=0,$ or
\be \la{azxx} 4\beta^2=\frac{6}{(n-1)(2n-1)}, \ \ \ \alpha^2=\frac{2n-4}{2n-1}. \ee
\vskip 0.2cm
\noindent 2) If the coefficients $Ch_2(Q)$, $Ch_4(Q),$  vanish, then either $\alpha=\pm 1,$ $\beta=0,$ or
\be \la{azxx+}
4{\beta}^2=\frac{12}{(n-1)(2n-1)}, \ \ \ {\alpha}^2=\frac{2n-7}{2n-1}. \ee
In particular, in both cases $\alpha^2< 1$ and $\beta \neq 0,$ unless $\alpha=\pm 1,$ $\beta=0.$
\el

\pr Set $n=\deg Q.$
By Lemma \ref{l1}, for some $b_0\in \R$ we have: $$C_i(Q)=C_i(F\circ T_s\circ (\alpha x+\beta))=C_i(b_0T_{n/s}\circ T_s\circ (\alpha x+\beta))=C_i(b_0T_n\circ (\alpha x+\beta)), \ \ \ 0\leq i \leq s-1,$$
implying that
$$Q=(b_0T_n)\circ (\alpha x+\beta)+R_2,$$ where $R_2$ ia a polynomial such that $\deg R_2\leq n-s$.
Thus, if $Ch_2(Q)=0,$ we have:
\be \la{0} Q=c_0T_n+c_1T_{n-1}+c_3T_{n-3}+c_4T_{n-4}+ R_{1}=b_0T_n\circ (\alpha x+\beta)+R_{2},\ee
where  $\deg R_1, \deg R_2\leq n-5$, and $c_0,c_1,c_3,c_4,b_0\in \R$.
Changing $x$ to $x/2$, and $R_i(x)$ to $2R_i(x/2)$, we obtain a similar equality
\be \la{4}  Q=c_0T_n^*+ c_1T_{n-1}^*+ c_3T_{n-3}^*+ c_4T_{n-4}^*+R_{1}={ b_0}T_n^{*}\circ (\t\alpha x+\t\beta)+R_{2},\ee
where $\t\beta=2\beta,$ $\t\alpha=\alpha.$ Furthermore, without loss of generality we may assume that $c_0=1,$ implying
that $ b_0=1/{\t \alpha}^n$ and $c_1=\t\beta n/\t\alpha$. Thus, we can rewrite \eqref{4} in the form
\be \la{5} Q=T_n^*+\frac{\t \beta n}{\t \alpha}T_{n-1}^*+c_3T_{n-3}^*+ c_4T_{n-4}^*+ R_{1}=\frac{1}{\t \alpha^n}T_n^{*}\circ (\t\alpha x+\t\beta)+ R_{2}.\ee

Calculating $C_2(Q),$ $C_3(Q),$ $C_4(Q)$ for both representations of $Q$ in \eqref{5} using
formula \eqref{f} (and the Taylor formula, for the second representation), we obtain the equalities
$$-n=\frac{1}{(n-2)!{\t \alpha}^2}\left[\frac{n!{\t \beta^2}}{2!}-n(n-2)!\right],$$
$$\frac{-{\t \beta}n(n-1)}{{\t \alpha}}+ c_3=\frac{1}{(n-3)!{\t\alpha^3}}\left[\frac{n!{\t \beta^3}}{3!}-n(n-2)!\t \beta\right],$$
$$\frac{n(n-3)}{2}+c_4=\frac{1}{(n-4)!{\t\alpha}^4}\left[
\frac{n!{\t \beta}^4}{4!}-\frac{n(n-2)!{\t \beta}^2}{2!}+\frac{n(n-3)(n-4)!}{2}\right].$$

It follows from the first of these equalities that
\be \la{gopa} {\t \alpha}^2=1-\frac{(n-1){\t \beta}^2}{2}.\ee If $c_3=0$, then
substituting this value of ${\t \alpha}^2$ into third equation we
obtain that either $\beta=0$ and then $\alpha=\pm 1$ by \eqref{gopa}, or \eqref{azxx} holds.
Similarly, if $c_4=0$, then
substituting the value of ${\t \alpha}^2$ from \eqref{gopa} into the fourth equation we
obtain
$$\frac{n!{\t \beta}^4}{4!}-\frac{n(n-2)!{\t \beta}^2}{2!}+\frac{n(n-3)(n-4)!}{2}= \frac{(n-4)!n(n-3)}{2}\left[\frac{(n-1)^2{\t \beta}^4}{4}-(n-1){\t \beta}^2+1\right]
$$ implying that, either $\t \alpha=\pm 1,$ $\t \beta=0,$  or  \eqref{azxx+} holds.

Finally, clearly $\alpha^2< 1$, $\beta \neq 0,$ unless $\alpha=\pm 1,$ $\beta=0.$
\qed

\bl \la{dfg3} Let $Q\in C(n,m_1,m_2)$, where $m_1$ and $m_2$ are
coprime, $m_1\geq 3,$ $m_2\geq 3$. Then at least one of the coefficients $Ch_2(Q),$ $Ch_4(Q),$ $Ch_6(Q)$ vanishes.
\el
\pr Assume that $Ch_2(Q)\neq 0,$ $Ch_4(Q)\neq 0,$ and show that this implies that $Ch_6(Q)=0.$ First, $Ch_2(Q)\neq 0$ implies by \eqref{inv} that $m_2\mid n-2.$
It follows now from $Ch_4(Q)\neq 0$ by \eqref{inv} that $m_1=4.$
Therefore, $Ch_6(Q)=0$ for otherwise \eqref{inv} implies that $m_2=4$ in contradiction with $\GCD(m_1,m_2)=1.$ \qed

\bl \la{ll3} Let $Q=U\circ T_s\circ \alpha x$, where $U\in \R[x],$ $\alpha \in \R\setminus\{0\},$ and $s\geq 6.$
\vskip 0.2cm
\noindent 1) If the coefficient $Ch_2(Q)$ vanishes, then $\alpha^2=1.$
\vskip 0.2cm
\noindent 2) If the coefficient $Ch_4(Q)$ vanishes, then
either $\alpha^2=1,$ or $\alpha^2=\frac{n-3}{n-1}.$
\vskip 0.2cm
\noindent 3) If the coefficient $Ch_6(Q)$ vanishes, then $\alpha^2$ is a root of the equation
\be \la{kon} (n^2-3n+2)t^2+(-2n^2+12n-16)t+(n^2-9n+20)=0.\ee In particular, in all the cases the inequality $\alpha^2< 1$ holds, unless $\alpha^2= 1$.
\el

\pr Set $n=\deg Q.$ As in Lemma \ref{l4} we arrive to the equality
\be \la{00} T_n^{*}+c_2T_{n-2}^{*}+c_4T_{n-4}^{*} + c_6T_{n-6}^{*}+ R_1=\frac{1}{\t \alpha^n}T_n^{*}\circ (\alpha x)+ R_2,\ee
where $\deg R_1, \deg R_2\leq n-7$ (since obviously $c_1=c_3=c_5=0$).
Calculating now $C_2(Q),$ $C_4(Q),$ $C_6(Q)$ for both representations of $Q$ in \eqref{00}, we obtain:
$$-n+ c_2=-\frac{n}{\alpha^{2}},$$
$$\frac{n(n-3)}{2}-(n-2)c_2+c_4=\frac{n(n-3)}{2\alpha^4},$$
$$-\frac{n(n-4)(n-5)}{6}+\frac{(n-2)(n-5)c_2}{2}-(n-4)c_4+c_6=-\frac{n(n-4)(n-5)}{6\alpha^6}$$

It follows from the first equality that $$c_2=\frac{n(\alpha^2-1)}{\alpha^2},$$ implying that
if $c_2=0$, then $\alpha^2=1.$ Substituting now the value of $c_2$ into the second equality
we obtain that
$$c_4=\frac{n(n-3)}{2\alpha^4}-\frac{n(n-3)}{2}+\frac{n(n-2)(\alpha^2-1)}{\alpha^2}=n\left(\frac{(n-1)\alpha^4-2(n-2)\alpha^2+(n-3)}{2\alpha^4}\right).
$$ Since $$(n-1)\alpha^4-2(n-2)\alpha^2+(n-3)=(n-1)(\alpha^2-1)\left(\alpha^2-\frac{n-3}{n-1}\right),$$
this implies that if $c_4=0$, then either $\alpha^2=1,$ or $\alpha^2=\frac{n-3}{n-1}.$

Finally, $$c_6=-\frac{n(n-4)(n-5)}{6\alpha^6}+\frac{n(n-4)(n-5)}{6}-\frac{(n-2)(n-5)c_2}{2}+(n-4)c_4=$$
$$=-\frac{n(n-4)(n-5)}{6\alpha^6}+\frac{n(n-4)(n-5)}{6}-\frac{(n-2)(n-5)}{2}\frac{n(\alpha^2-1)}{\alpha^2}+$$
$$+(n-4)n\left(\frac{(n-1)\alpha^4-2(n-2)\alpha^2+(n-3)}{2\alpha^4}\right)=$$
$$
=n\left(\frac{(n^2-3n+2)\alpha^6+(-3n^2+15n-18)\alpha^4+(3n^2-21n+36)\alpha^2-n^2+9n-20}{6\alpha^6}\right).
$$
Now the statement follows from the factorization
$$
(n^2-3n+2)\alpha^6+(-3n^2+15n-18)\alpha^4+(3n^2-21n+36)\alpha^2-n^2+9n-20=$$ $$=(\alpha^2-1)\left((n^2-3n+2)\alpha^4+(-2n^2+12n-16)\alpha^2+(n^2-9n+20)\right). $$

In order to finish the proof we only must show that the absolute values of roots of
equation \eqref{kon} are less than one. Solving \eqref{kon}, we find two roots
$$t_1=\frac{n^2-6n+8-\sqrt{3n^2-18n+24}}{n^2-3n+2}=\frac{\sqrt{(n-2)(n-4)}(\sqrt{(n-2)(n-4)}-\sqrt{3})}{(n-1)(n-2)},
$$
$$t_2= \frac{n^2-6n+8+\sqrt{3n^2-18n+24}}{n^2-3n+2}=\frac{\sqrt{(n-2)(n-4)}(\sqrt{(n-2)(n-4)}+\sqrt{3})}{(n-1)(n-2)}.$$
Clearly, for $n\geq 6$ the inequality $0<t_1<t_2$ holds. Finally,
$$t_2-1=\frac{\sqrt{3(n-2)(n-4)}-3n+6}{(n-1)(n-2)}=\frac{\sqrt{3(n-2)}(\sqrt{n-4}-\sqrt{3(n-2)})}{(n-1)(n-2)}<0. \ \ \ \Box$$

\vskip 0.2cm
\noindent{\it Proof of Proposition  \ref{vot}.}
Assume first that $Ch_1(Q)\neq 0.$
Then Lemma \ref{dfg} implies that either $Ch_2(Q)$, $Ch_3(Q)$ vanish, or $Ch_2(Q)$, $Ch_4(Q)$ vanish.
It follows now from Lemma \ref{l4} that, unless $\alpha=\pm 1,$ $\beta =0,$ the conditions $\alpha<1$, $\beta\neq 0$ hold. So, assume that $\alpha<1$, $\beta\neq 0$.

Rewrite equality \eqref{uraba2} in the form
\be \la{uraba3} P=\t V_1\circ T_{\t m_1}+\t V_2\circ T_{\t m_2}=U\circ T_{m_1m_2}\circ \left(\frac{x-\beta}{\alpha}\right).\ee Since $\beta \neq 0,$ Corollary \ref{c1}
applied to \eqref{uraba3} implies that $C_1(P)\neq 0.$ Applying now Lemma \ref{dfg} and Lemma \ref{l4} to \eqref{uraba3} in the same way as before to equality \eqref{uraba}, we  conclude that
$1/\alpha<1.$ The contradiction obtained proves that $\alpha=\pm 1,$ $\beta =0.$

Assume now that $Ch_1(Q)= 0.$ Then $\beta=0$, by Corollary \ref{c1}.
Furthermore, by Lemma \ref{dfg3}
at least one of the coefficients $Ch_2(Q),$ $Ch_4(Q),$ $Ch_6(Q)$ vanish,
implying by Lemma \ref{ll3} that, unless $\alpha=\pm 1,$ $\beta =0,$ the condition $\alpha<1$ holds. Since $\beta=0$ implies by Corollary \ref{c1} that $C_1(P)=0$ in view of \eqref{uraba3}, the assumption $\alpha<1$  leads to a contradiction in the same way as before. \qed

\subsection{Proof of Theorem \ref{t2} in the cases (3,1), (3,2), (3,3).}
The  case (3,1) reduces  to the case (2,1) as follows.
We start from the equality
\be \la{ura} Q=V_1\circ T_{2m_1}+V_2\circ T_{2m_2}+V_3\circ xR(x^2)\circ T_{m_1m_2}
=\t U\circ
 x^2{\t R}^2(x^2)\circ(\alpha x+\beta),\ee where $V_1,V_2,V_3,R,\t R, \t U\in \R[x]$, $\alpha,\beta\in \R,$ $\alpha\neq 0,$
and $m_1\geq 3,$ $m_2\geq 3$ are coprime and odd.
It follows from the first representation for $Q$ in \eqref{ura} that $Q$ can be written in the form
\be \la{ebi} Q=d_nT_n+d_{n-1}T_{n-1}+\dots +d_{1}T_1+d_0,\ \ \ d_i\in \R,\ee
where $d_i=0$, unless $i$ is divisible either by $2m_1,$ or by $2m_2$, or by $m_1m_2$.
Clearly, conditions imposed on $m_1,m_2$ imply that
\be \la{dif1} \vert 2m_1k_1-2m_2k_2\vert\geq 2, \ \ \ {\rm unless} \ \ \ \vert 2m_1k_1-2m_2k_2\vert=0,\ee
\be \la{dif1+} \vert 2m_ik_1-2m_ik_2\vert\geq  2m_i\geq  6, \ \ \ {\rm unless} \ \ \ \vert 2m_ik_1-2m_ik_2\vert=0,\ \ \ i=1,2\ee
\be \la{dif2} \vert 2m_ik_1-m_1m_2k_2\vert\geq  m_i\geq 3, \ \ \ {\rm unless} \ \ \ \vert 2m_ik_1-m_1m_2k_2\vert=0,\ \ \ i=1,2\ee
\be \la{dif3} \vert m_1m_2k_1-m_1m_2k_2\vert\geq  m_1m_2 \geq 15, \ \ \ {\rm unless} \ \ \ \vert m_1m_2k_1-m_1m_2k_2\vert=0.\ee
Therefore, $Ch_1(Q)=0$, implying that $C_1(Q)=0$, since $C_1(T_n)=0.$
It follows now from the second representation for $Q$ in \eqref{ura} by Corollary \ref{c1} that $\beta=0$.
Since the polynomial $\t W_1=z^2\circ (\alpha x+\beta)$ satisfies $\t W_1(a)=\t W_1(b)$, this implies that
$a=- b.$ Therefore, solution $P,Q$ also belongs to the case (2,1) considered earlier (see the remarks after Theorem \ref{mpc}).

In the case (3,2)
there exist  $V_1,V_2,V_3, U, R, \t V_1, \t V_2, \t U\in \R[x]$ and $\alpha,\beta\in \R$, $\alpha\neq 0,$  such that
\be \la{ura+} Q=V_1\circ T_{2m_1}+V_2\circ T_{2m_2}+V_3\circ xR(x^2)\circ T_{m_1m_2}
=\t U\circ T_{\t m_1\t m_2} \circ (\alpha x+\beta),\ee
\be \la{rau+} P=\t V_1\circ T_{\t m_1} \circ  (\alpha x+\beta)+\t V_2\circ T_{\t m_2}\circ  (\alpha x+\beta)=U\circ x^2R^2(x^2)\circ T_{m_1m_2}\ee
where $m_1\geq 3$, $m_2\geq 3$ are odd, $\GCD(m_1,m_2)=1,$ and $\t m_1\geq 2$, $\t m_2\geq 2$, $\GCD(\t m_1,\t m_2)=1,$
Further, without loss of generality we may assume that $a\neq - b$, for otherwise $P,Q$ belongs to the case $(2,2).$ Besides, we may assume that $\t m_1\geq 3$, $\t m_2\geq 3,$ for otherwise $P,Q$ belongs to the case $(3,1).$

Since equalities \eqref{ura+}, \eqref{rau+} may be written in the form
$$ Q=(V_1\circ T_2+ V_3\circ xR(x^2)\circ T_{m_2})\circ T_{m_1} +(V_2\circ T_2)\circ T_{m_2}
=\t U\circ T_{\t m_1\t m_2} \circ (\alpha x+\beta),$$
$$ P=\t V_1\circ T_{\t m_1} \circ  (\alpha x+\beta)+\t V_2\circ T_{\t m_2}\circ  (\alpha x+\beta)=(U\circ x^2R^2(x^2))\circ T_{m_1m_2},$$
it follows from Proposition \ref{vot} that $\alpha=\pm 1,$ $\beta =0.$
Since we assumed that $a\neq - b,$ it follows now from the equalities
$$T_{2m_1}(a)=T_{2m_1}(b), \ \ \ T_{2m_2}(a)=T_{2m_2}(b), \ \ \ T_{{\t m}_1}(\pm a)=T_{{\t m}_1}(\pm b)$$
by  Lemma \ref{skun}, b), taking into account \eqref{kots}, that
$T_s(a)=T_s(b)$, where either $s=\GCD(2m_1, {\t m}_1)$, or $s=\GCD(2m_2, {\t m}_2).$ Finally,
since
$$Q=\t U\circ T_{\t m\t n}\circ (\pm z)=\t U\circ (\pm  T_{\t m\t n})=\t U\circ (\pm  T_{\t m\t n/s})\circ T_s$$
and
$$P=U\circ x^{2}R^2(x^{2})\circ  T_{m_1m_2}=U\circ x^{2}R^2(x^{2})\circ  T_{m_1m_2/s}\circ T_s,$$
we conclude that \eqref{c} holds for $W=T_s.$

The proof in the case (3,3) is similar:
there exist  $V_1,V_2,V_3, U, R, \t V_1, \t V_2, \t V_3, \t U, \t R\in \R[x]$ and $\alpha,\beta\in \R$, $\alpha\neq 0,$  such that
$$ Q=V_1\circ T_{2m_1}+V_2\circ T_{2m_2}+V_3\circ xR(x^2)\circ T_{m_1m_2}
=\t U\circ x^2\t R^2(x^2)\circ T_{\t m_1\t m_2} \circ (\alpha x+\beta),$$
$$ P=\t V_1\circ T_{\t 2m_1} \circ  (\alpha x+\beta)+\t V_2\circ T_{\t 2m_2}\circ  (\alpha x+\beta)+\t V_3\circ x\t R(x^2)\circ T_{\t m_1\t m_2}
=U\circ x^2R^2(x^2)\circ T_{m_1m_2}$$
where $m_1\geq 3$, $m_2\geq 3$, $\t m_1\geq 3$, $\t m_2\geq 3$ are odd, $\GCD(m_1,m_2)=1,$ $\GCD(\t m_1,\t m_2)=1.$ Moreover,
without loss of generality we may assume that $a\neq - b$.
Further, using  Proposition \ref{vot} we conclude as above that  $\alpha=\pm 1,$ $\beta =0.$  Finally, it follows from
the equalities
$$T_{2m_1}(a)=T_{2m_1}(b), \ \ \ T_{2m_2}(a)=T_{2m_2}(b), \ \ \ T_{{\t 2m}_1}(\pm a)=T_{{\t 2m}_1}(\pm b)$$
that \eqref{c} holds for $W=T_s,$  where either $s=\GCD(2m_1, {2\t m}_1)$, or $s=\GCD(2m_2, 2{\t m}_2).$

redpajama_set_name: RedPajamaArXiv


text:
Majorca with kids: a holiday that will please everyone
Majorca isn't just for beach bums, this Balearic Island offers up plenty of fun for the whole family. Check out our guide on enjoying Majorca with kids.
There are a staggering 262 beaches to enjoy in Majorca, the largest of the Balearic Islands. And it's Majorca's gorgeous coastline, azure blue seas, hilltop towns and culturally magnificent cities that make it a fantastic family destination.
Explore the pine-fringed coves of Cala d'or and find a secluded spot to paddle in the cool shallow waters of Cala Serena, then take the kids to wander to the marina and go yacht spotting. Get lost in the Old Town of Palma and grab a creamy sweet ensaimada from a traditional bakery and then head to the waterfront to marvel at the majestic cathedral.
From secluded beaches to golden-stoned villages, bike rides, watersports, hiking, and plenty of child-friendly attractions, Majorca for families is a must. The only trouble is, trying to decide where to head first and whether to tear yourselves away from the poolside fun.
Best places in Majorca for families
Beach-loving families will love Playa de Muro. Small children will be happy splashing in the shallow turquoise sea, while teens will love kiteboarding and the watersports on offer. Meanwhile, mum and dad can sit back on the comfy chairs at the beachfront bar and restaurant, Royal Beach.
If you're not wild about the beach, there is plenty more things to do with the kids in Majorca.
Hotels - Palma de Majorca
Best family resorts in Majorca
From Puerto de Alcudia and Cala Millor to Soller, there really is so much choice when it comes to the best places for families in Majorca. Which is why we've rounded up a top pick of fantastic family-friendly Majorca hotels.
Superb facilities close to Playa de Muro
Located to the stunning shores of Playa de Muro which is just a short five-minute stroll away, the Hotel VIVA Blue & Spa is a firm favourite with both families and sports enthusiasts. Kids will adore the children's club which entertains from day until night with a variety of activities, as well as the pirate-themed splash park with slides. Older kids and fitness fanatics won't be able to get enough of the state-of-the-art training facilities including cycling workshops and a multi-sports court. The hotel is also close to the impressive landscapes of Albufera Natural Park.
The premium apartments are spacious and come with a private terrace and jacuzzi. And with three large pools, one of which is for adults only, as well as a Balneum Spa which offers up tempting massage treatments and a Turkish Bath, parents can sunbathe on the comfortable Bali beds and enjoy a long lunch and great views on the outdoor terrace at Restaurant Mediterraneo.
VIVA Blue & Spa
Playa de Muro
8.7 Excellent (2262 reviews)
Outstanding luxury with an acclaimed restaurant
For those parents who want to combine a deluxe retreat with a family holiday then the five-star Belmond La Residencia is the perfect choice. The sublime setting between the mountains and the sea, the 16th century Belmond is situated in the picturesque village of Deia and is close to a pretty pebbled beach.
Rather than feeling like they're a nuisance, children are welcomed with open arms. The Smile Club is open to 4-12 year-olds, and there are excursions including donkey rides, cooking and art classes, tennis coaching and play stations, Ipads and toys can be borrowed. Meanwhile, with babysitting services available, parents can have date night, every night – and dine in the intimate El Olivo restaurant which is considered one of the finest restaurants on the island, followed by live music and drinks on the pretty terrace of Café Miró. With a spa, two outdoor pools, indoor pools and sauna, there are also many areas to kick back and relax.
La Residencia, A Belmond Hotel, Mallorca
9.2 Excellent (292 reviews)
Sleek and contemporary cool in Alcudia
The Zafiro Palace Alcudia occupies is very own private space in the Port of Alcudia. It's a large, contemporary resort, and its modern lines and glass facade gives it a light, airy feel, keeping in tune with the surroundings. Family suites are generous in size, with two bedrooms and direct access to the pools and Balinese beds.
Little kids will be hard pushed to leave the bubble pool, pirates ship and slides, but they might be persuaded to go to the mini club which has a bouncy castle and play area, as well as the superb facilities catering to kids 4-15 years of age. From an abundance of sporting activities, even indoor cycling, and local demonstrations from bird shows, boredom won't ever be a problem.
Parents will want to head to the Zen Spa or lounge by the 12 pools – one of which is adults-only. Meanwhile, there is a choice of five restaurants where you can sample Majorcan delicacies or a taste of the East at the sushi bar-restaurant. The Verandah buffet restaurant has numerous food stations which will entice even tiny palettes, particularly the ice-cream on tap throughout the day.
Puerto de Alcudia
A four-star all-inclusive in Pollenca Bay
With superb views over the pine-covered bay of Pollenca, the PortBlue Club Pollentia Resort & Spa is well-placed to explore the gorgeous stretch of Playa Can Cap Bou Breach as well as the nearby town. However, it's the epic pool scene that will prove popular with both young and old guests.
You'll find plenty of space to lounge by the main pool, and keep an eye on little ones in the splash area. There is even an adults-only section with comfy day beds. Get some alone time, too, at the Ocean Spa & Wellness, where you can book yourself a personalised massage treatment.
Roomy family suites have direct access to the gardens and lakes, while the food at PortBlue is excellent with a good mix of self-service and a la carte restaurants as well as three bars to enjoy evening cocktails. For children there is a creche for kids under 4 and a club for children up to the age of 12, where they can enjoy a playground, sporting activities and water play.
PortBlue Club Pollentia Resort & Spa
8.4 Very good (5611 reviews)
Superior seaside hotel close to Playa de Palma
With a great mix of contemporary features and a modern vibe, The Iberostar Playa de Palma has a sleek white interior contrasted with bursts of turquoise. From the incredible rooftop infinity pool to Spa Sensations which are for adults only, to first-class all-inclusive dining there are plenty of secluded spots for parents to really kick back. For those who want the ultimate space to drink in the views of the Bay of Palma, book into the Star Prestige suites, where the whole family can enjoy a private hydromassage tub out on a private terrace.
While there are no onsite facilities directed at kids, no matter. Just nip next door to the Iberostar Cristina and get the best of both worlds. Kids will love the jungle world and Star Camp as well as the kids' pool. It's the hotel's close proximity to the Palma though which means you get the best of both worlds – a chance to relax by the coast, but the excitement of all the fun activities in the city for kids, too.
Iberostar Selection Playa de Palma
Five-star decadence with panoramic views
Standing on the cliff top and overlooking the stunning coast of Port de Soller and the Tramuntana mountains is the uber-luxe Jumeirah Port Soller. Built to blend in with its hilltop surroundings, guests will be instantly impressed by the lobby's giant snail chandelier. Interiors are relaxed and neutral throughout including the well-appointed rooms all with mountain views.
While there is a separate family pool, mum and dad can sneak off to the infinity pool and drink in the sweeping vistas. The two Hammams, large glass-fronted sauna and hydro pool at the Talise Spa also make use of these magnificent views, but perhaps you might tear yourself away for the Five Senses massage, which includes a coma-inducing foot scrub.
Kids will adore the Barbaroja's Kids Club, which is kitted out with entertaining features, as well as organised activities including a fantastic nature trail with a guide and movie nights. Babysitting services means parents can also enjoy freshly prepared seafood at the romantic and elegant Cap Roig restaurant and finish off their evening with a drink at the Sunset Sushi Lounge which has calming seascapes.
Puerto de Sóller

redpajama_set_name: RedPajamaCommonCrawl


text:
using System;
using CertificateServices.Enumerations;
using CertificateServices;
using System.Collections.Generic;
using CertificateManager.Entities.Attributes;
using CertificateManager.Entities.Interfaces;
using CertificateManager.Entities.Extensions;

namespace CertificateManager.Entities
{
    [Repository("Certificate")]
    public class GetCertificateEntity : ILoggableEntity
    {
        public Guid Id { get; set; }
        public string DisplayName { get; set; }
        public string Thumbprint { get; set; }
        public HashAlgorithm HashAlgorithm { get; set; }
        public CipherAlgorithm CipherAlgorithm { get; set; }
        public WindowsApi WindowsApi { get; set; }
        public KeyUsage KeyUsage { get; set; }
        public int KeySize { get; set; }
        public DateTime ValidTo { get; set; }
        public DateTime ValidFrom { get; set; }
        public CertificateStorageFormat CertificateStorageFormat { get; set; }
        public CertificateSubject Subject { get; set; }
        public bool HasPrivateKey { get; set; }
        public List<AccessControlEntry> Acl { get; set; }

        public string GetDescription()
        {
            return string.Format("Certificate viewed thumbprint: {0};id: {1};displayname: {2}", Thumbprint, Id, DisplayName);
        }

        public string GetId()
        {
            return Id.GetId();
        }
    }
}

redpajama_set_name: RedPajamaGithub


text:
Honolulu Mayor Rick Blangiardi signs measure to incentivize affordable housing
By Ashley Mizuo amizuo@staradvertiser.com
STAR-ADVERTISER / APRIL 5
Rick Blangiardi
Honolulu Mayor Rick Blangiardi has signed into law a measure that aims to increase affordable rental housing by offering financial grants to private developers. Read more
Honolulu Mayor Rick Blangiardi has signed into law a measure that aims to increase affordable rental housing by offering financial grants to private developers.
The financial grants would be distributed to developers who rent to households earning below the annual area median income.
"The building up of affordable rentals, more so than ever before, is of extreme importance," Blangiardi said Thursday.
"We have a shortage, but we have an even greater need. And we need to match that up."
In Hawaii the area median income for one person is $71,100 and for a family of four is $101,600, according to the Hawaii Community Development Authority.
For developments renting to households above 60% of the average median income, the incentive would be $11.25 per square foot of living space in the building permit but would not exceed $9,000.
For developments renting to households below 60% of the average median income, the incentive would be $15 per square foot of living space in the building permit but would not exceed $15,000.
Although during the news conference Blangiardi said the housing must stay affordable permanently, an ordinance from last year limits it to 15 years.
The bill also limits the total amount of grant money offered from the program to $10 million.
Bill 1 is a continuation of another five-year affordable-housing pilot program that began in 2019 which offers incentives such as a 10-year property tax waiver and wastewater connections for affordable-housing developments.
Island Business Management's Ryan Tanaka, who worked closely with Blangiardi and the Honolulu City Council on the measure, is working with clients to develop affordable housing based on those initiatives.
"We expect from Bill 1 a total of 30 buildings to develop over the next three years. And as (the) mayor has said, this will contribute 1,100 affordable housing units," he said.
"Despite the significant development risk, the razor-thin margins … the escalating construction costs — we're not putting luxury condominiums, but these are well-designed buildings. Each unit will have a full kitchen, full bath, a built-in washer-dryer, air conditioning and modern finishes."
Eleven hundred affordable housing units would make a dent in the tens of thousands of new units needed to house Hawaii's projected population, according to a 2019 report done by the Honolulu Department of Planning and Permitting.
Blangiardi explained that this measure was just the start to incentivizing affordable housing.
"We're looking at a whole lot of different projects," he said.
"We're looking at the notion of Oahu where people live in communities. We're looking across the board at a number of other projects for building affordable."
Correction: An earlier version of this story said the developments built under the new ordinance would be permanently for affordable housing. However, an ordinance signed last year establishes a 15-year minimum.
Rearview: Sears, Ala Moana and peanuts  — an unlikely combination
Shark protection bill awaits Hawaii Gov. David Ige's signature


text:
using Quartz;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Quartz.TextToSchedule
{
    /// <summary>
    /// The results of a <see cref="ITextToSchedule.Parse"/>.
    /// You can register the given schedule with methods defined in the class.
    /// </summary>
    public class TextToScheduleResults
    {
        /// <summary>
        /// Gets the groups of paired Triggers and Calendars.
        /// </summary>
        public List<RegisterGroup> RegisterGroups { get; private set; }

        /// <summary>
        /// Initializes a new instance of the <see cref="TextToScheduleResults" /> class.
        /// </summary>
        public TextToScheduleResults()
        {
            RegisterGroups = new List<RegisterGroup>();
        }

        /// <summary>
        /// Adds the specified trigger builder.
        /// </summary>
        /// <param name="triggerBuilder">The trigger builder.</param>
        public void Add(TriggerBuilder triggerBuilder)
        {
            RegisterGroups.Add(new RegisterGroup(triggerBuilder));
        }

        /// <summary>
        /// Adds the specified trigger and calendar as a pair.
        /// </summary>
        /// <param name="trigger">The trigger.</param>
        /// <param name="calendar">The calendar.</param>
        public void Add(TriggerBuilder trigger, ICalendar calendar)
        {
            RegisterGroups.Add(new RegisterGroup(trigger, calendar));
        }

        #region Registering With Scheduler

        /// <summary>
        /// Registers the calendars to the scheduler.
        /// </summary>
        /// <param name="sched">The sched.</param>
        private void RegisterCalendars(IScheduler sched)
        {
            foreach (var g in RegisterGroups)
            {
                ICalendar cal = g.Calendar;

                if (cal != null)
                {
                    string newCalendarName = string.Format("CAL_{0}", Guid.NewGuid());
                    sched.AddCalendar(newCalendarName, cal, true, true);

                    g.TriggerBuilder.ModifiedByCalendar(newCalendarName);
                }
            }
        }

        /// <summary>
        /// Schedules the Triggers with a given job key.
        /// </summary>
        /// <param name="sched">The sched.</param>
        /// <param name="jobKey">The job key.</param>
        public void ScheduleWithJobKey(IScheduler sched, JobKey jobKey)
        {
            RegisterCalendars(sched);

            foreach (var group in RegisterGroups)
            {
                group.TriggerBuilder.ForJob(jobKey);

                ITrigger trigger = group.TriggerBuilder.Build();
                sched.ScheduleJob(trigger);
            }
        }

        /// <summary>
        /// Schedules the with the provided <see cref="IJobDetail"/>.
        /// </summary>
        /// <param name="sched">The sched.</param>
        /// <param name="jobDetail">The job detail.</param>
        public void ScheduleWithJob(IScheduler sched, IJobDetail jobDetail)
        {
            RegisterCalendars(sched);

            if (RegisterGroups.Count > 1) // use the bulk add method if using more than 1
            {
                List<ITrigger> triggers = new List<ITrigger>();
                RegisterGroups.ForEach(x => triggers.Add(x.TriggerBuilder.Build()));

                var dic = new Dictionary<Quartz.IJobDetail, Quartz.Collection.ISet<ITrigger>>();
                Quartz.Collection.ISet<ITrigger> set = new Collection.HashSet<ITrigger>(triggers);
                dic.Add(jobDetail, set);

                sched.ScheduleJobs(dic, true);
            }
            else if (RegisterGroups.Count == 1)
            {
                ITrigger trigger = RegisterGroups[0].TriggerBuilder.Build();
                sched.ScheduleJob(jobDetail, trigger);
            }
        }

        #endregion
    }

    /// <summary>
    /// A Trigger and Calendar pair.
    /// </summary>
    public class RegisterGroup
    {
        public TriggerBuilder TriggerBuilder { get; private set; }
        public ICalendar Calendar { get; private set; }

        public RegisterGroup(TriggerBuilder trigger)
        {
            TriggerBuilder = trigger;
        }
        public RegisterGroup(TriggerBuilder trigger, ICalendar calendar)
        {
            TriggerBuilder = trigger;
            Calendar = calendar;
        }
    }
}


If you're struggling to find the right size cutlery to eat very small food items then the Skoowi Mini necklace is the one for you. Featuring a teeny knife, fork and spoon; it's all you need for pea sized portions!
Suspended from a fine trace chain, choose from 16", 18", 20" and 22".
The knife, fork and spoon measure 1.5cm in length. Tiny food items recommended.


This Salvaged garment is a 100% vintage 1990s black sheer blouse with mother-of-pearl buttons down front and beautiful embroidered floral beaded detailing on sleeves and cuffs. CONDITION - Excellent vintage condition, may be some natural signs of wear due to age. SIZE - Model is a size 8 and 5'7". Garment size 6-10 dependent on fit desired. DELIVERY - Next day delivery available Monday-Thursday on orders made before 1pm. Orders made on Friday will be posted on Monday.


You are at:Home»Data»Where Were the Most Jobs Created in 2018?
Which states created the most jobs and which ones lost them? That question was recently put to the test by the NAHB's Eye on Housing blog. Using data from the Bureau of Labor Statistics total non-farm employment in December, 2018 increased by 2.6 million jobs, compared with a gain of 2.2 million in 2017. It turns out that the South created the most number of jobs by adding 1.2 million total non-farm employment to the area, while the Northeast created the least number of non-farm jobs. The bottom-line; just look at the map and follow the money.


text:
La Aldea de La Fuente del Conde, situada al sur del término municipal de Iznájar, al pie de la sierra de Campoagro, es también conocida popularmente  como "La Alcubilla" o "La Arcuilla"
Su nombre se debe al conde de la Revilla, que vivía en la Aldea y que era propietario de una fuente con cuatro abundantes caños de agua, así como de la mayoría de las tierras sureñas de Iznájar.
Coronando la Aldea se encuentra un magnífico mirador, la "Cruz de Magán", con una pequeña ermita cuyo origen se relaciona con un antiguo santuario ibérico que se ubicase en ese mismo lugar.
Jalonando el cerro hay numerosas casas cueva que estuvieron habitadas durante el Neolítico y en épocas posteriores. También hay constancia de asentamientos de poblados argáricos, ibéricos y romanos, habiéndose encontrado numerosos restos arqueológicos en las cercanías que así lo confirmarían, como falcatas, puntas de flecha, lanzas, regatones y fragmentos de cerámicas ibéricas.
Elementos arquitectónicos interesantes en Fuente del Conde son su Iglesia, una moderna construcción iniciada hacia 1985 y finalizada en mayo de 1994, sus lavaderos y fuentes públicas, como la "Fuente de Magán", la "Fuente de la Alamea" o la "Fuente de la Alcubilla". Destacan también algunas encinas que, por su tamaño, formas y edad son considerados árboles singulares del municipio.
Celebra sus fiestas en honor de la Santa Cruz, del 1 al 3 de mayo, con una típica y ancestral romería.

Iznájar

redpajama_set_name: RedPajamaWikipedia


Villa Tanqueta has 3 bedrooms distributed in a double bedroom, a single bedroom and a double bedroom, a large dressing room with blankets and towels, two bathrooms, air conditioning, heating and internet service FREE WI-FI, kitchen with vitro, microwave, Sandwich maker, toaster, juice squeezer, electric coffee maker, crockery for 6 people. Living room with tv with satellite and foreign channels, stereo, safe.
Rooms fully equipped and comfortable. They have all the comforts of a superior hotel. High quality beds and pillows to guarantee your rest.


text:
Przedsionkowy peptyd natriuretyczny, peptyd natriuretyczny typu A, ANP (od ang. atrial natriuretic peptide lub A-type natriuretic peptide) – hormon peptydowy wytwarzany przez ściany przedsionka serca pod wpływem wysokiego stężenia jonów sodu, dużej ilości płynu pozakomórkowego lub dużej ilości krwi.

Peptyd ten hamuje resorpcję zwrotną jonów sodu i wody głównie w kanalikach zbiorczych nerki i prowadzi do ich zwiększonego wydalania z moczem. Wpływa również na rozszerzanie i zwężanie pewnych naczyń krwionośnych (tętniczek doprowadzających i odprowadzających kłębuszków nerkowych), co wpływa na szybkość filtrowania płynów w nerkach, a to powoduje przyspieszenie produkcji moczu. Hamuje on układ renina–angiotensyna–aldosteron, poprzez stymulację syntezy prostaglandyn, oraz zmniejsza wydzielanie hormonu antydiuretycznego – przeciwdziała więc mechanizmom nasilającym niewydolność krążenia. Jego rola jest jednak niewielka, gdyż liczba i wrażliwość receptorów jest zmniejszona – mechanizm . Gen dla przedsionkowego peptydu natriuretycznego ma locus na 1p36.21.

Oznaczanie stężenia peptydów natriuretycznych w surowicy może być pomocne w diagnostyce niewydolności serca.

Przypisy

Hormony peptydowe
Geny na chromosomie 1


What we all need to do differently in developing our marketing plan an why.
How we can successfully roll out new techniques to our teams.
How we can measure our results in this overly analytical marketing world.
What "basics" that we have all done, are still working, and how to "glam" them up for better traction and results.
What to do if it seems nothing is working with your marketing efforts.
Who you can partner with for success- business, apartment marketing outlets and even your postman!!
Bio: For over 15 years, Stephanie Graves has shown her talent in dealing with property marketing and operations while offering a wide spectrum of insight into the multi-family world. She has spearheaded training and employee development programs that have increased operational efficiencies for property management companies nationwide Stephanie has a BA in marketing and is a frequent presenter for NAA, holds the CAPS and CAM designations and is a recognized NAA Education Faculty member.


If baroque defines a style that uses clearly defined decoration to create exuberance and grandeur, then there is no better way to describe or name this attractive, renaissance-styled vest then as the Gentleman's Baroque Vest. Made from quality brocade fabric, this vest is nothing if not grandiose in its design. The brocade is a baroque-styled damask pattern that displays floral and vine-like patterns that from afar create a visually complex yet eye-catching design. Being a vest, this garment is sleeveless, and in this garments particular case, collarless as well. When worn to the proper length the vest should be approximately hip-length. Real metal buttons run down the front of the vest and are used to fasten the vest closed (if you want to, that is), while the front also features two functional pockets. From the waist down, the vest cannot be buttoned and remains split open in the front. In the back, the split goes farther and remains open, allowing the garment to both breathe easily and allow for a wide range of movement. This jacket is also fully-lined with satin, making it exceedingly comfortable to wear. Here, the Gentleman's Baroque Vest is worn with a white collared shirt and knee breeches, which are shown for effect but not included. Combine the Gentleman's Baroque Vest with a tasteful, renaissance-styled shirt and a pair of elegant breeches or pants, and suddenly, you have a fully functional renaissance look that not only stylish, but also a touch noble and fancy, as well.
With golden colored lace trim along its collar, hem, lapel, shoulders, and cuffs, the Open Front Baroque Jacket is an example of refined Renaissance grace. This period jacket is made from antique velvet and fully lined for comfort.


text:
Market Square is a market square in Providence, Rhode Island. It is located at the intersection of present-day North Main Street and College Street at the base of College Hill. Market Square has historically functioned as a commercial, civic, and cultural locus of Providence.

History

Colonial era
In the 17th century the land comprising modern day Market Square was originally owned by Chad Brown, progenitor of the Brown family, later affiliated with Brown University.

The square's origin lies in a 1738 order by the Providence Town Council, which established a highway 123 feet in width, extending from Towne Street (now South Main Street) to the Great Bridge. At the time, the area was known as the Town Parade.

In 1744, a Haymarket was organized at the site, establishing the square as a local center of commerce.

The square's focal structure, Market House, was constructed between 1773 and 1775.

On March 2, 1775, Providence residents, inspired by the then-recent Boston Tea Party, gathered in Market Square to protest the Tea Act. The colonists heaped a large pile of English tea in the center of the square, covering the mound with a barrel of tar and lighting it ablaze.

Black history
As the commercial hub of colonial Providence, Market Square has been suggested as the likely site of slave sales, which constituted a significant portion of commerce in the 18th century city. The identification of the site as a slave market, however, has not been confirmed by primary sources. Historical references to slave sales in Providence suggest that these transactions traditionally occurred in private, commercial establishments rather than in a central, public space.

Enslaved black laborers contributed to the 1775 construction of Market House. Among these laborers was Pero Paget, a stonemason who also worked to built nearby University Hall at Brown University

In the 18th and 19th centuries, Market Square served as an important commercial venue for free Black entrepreneurs in the city.

19th century

In June 1843, President John Tyler toured New England while considering a potential third-party bid for re-election. One of Tyler's several Providence stops was at Market Square. The president enjoyed a meal at the Franklin House, an inn across from the Market House.

The 1847 construction of Union Station effectively shifted the city's commercial center to Exchange Place in Downtown Providence.

During the Civil War, Market Square was the site of several public "war meetings" presided by mayor Jabez C. Knight. During these meetings, public officials and dignitaries made public announcements and attempted to build up enlistment and support for the Union side during the war.

One such meeting was held in August 1862 to announce the Militia Act of 1862, which gave the state authority to draft. The act also allowed African-Americans to participate in the war as soldiers and war laborers. The announcement was received with "loud cheers." Governor William Sprague IV called for "colored citizens" to form a regiment, and promised to personally accompany this regiment into battle. In July 1863, after the Enrollment Act established a national draft, a blindfolded official selected names of conscripted men from a wheel in Market Square.

In 1882 the first electric arc lights in Providence were installed by the Rhode Island Electric Lighting Company on Market Square and Westminster Street.

On September 7 (or September 8), 1897, the anarchist and feminist Emma Goldman was arrested for "unlawful open air speaking" and "attracting a crowd" when she attempted to speak in at Market Square, during a four-month lecture tour. The mayor of Providence had warned Goldman that she would be arrested if she spoke in Providence. She had been traveling to lecture on topics such as "Why I am an Anarchist-Communist," "Woman", "Marriage", the recent assassination of the Spanish Premier, and a speech "Berkman's Unjust Sentence," about Alexander Berkman's imprisonment for the murder of Henry Clay Frick. After jailing Goldman overnight, the Providence authorities ordered her to leave town within 24 hours, or else face three months imprisonment.

RISD

Market House was acquired by Rhode Island School of Design (RISD) in 1948; Market Square is currently surrounded by the school's urban campus.

In 2016 Market Square was the site of a student organized protest 'Not Your Token' decrying racism and elitism on campus. Student organizers from Black Artists and Designers, created a list of mandates, which included increased hiring and retainment of faculty of color, faculty trainings and curricular changes. Protestors also demanded a memorial be erected in Market Square, acknowledging the site's purported associations with slavery and honoring victims of slavery in Rhode Island.

Gallery

See also

 Market House (Providence, Rhode Island)
 Kennedy Plaza

References

External links

Squares in Providence, Rhode Island


text:
Lo scudo di Enea è una delle armi possedute dall'eroe della mitologia greca e romana, protagonista dell'Eneide di Virgilio.

Storia
Mentre Enea si trova nella città del re arcade Evandro, la madre Venere si reca da Vulcano convincendolo a creare delle armi per il figlio. Enea, che aveva raggiunto Tarconte dietro consiglio di Evandro, viene raggiunto dalla madre, descritta come "splendente tra gli eterei nembi"; quella, dopo averlo abbracciato, consegna al figlio le armi perché possa sfidare in battaglia i Laurenti e Turno. Enea ammira l'elmo, la spada e la corazza, poi gli schinieri e la lancia, fino a soffermarsi sullo scudo. Vulcano gli incide sopra momenti della storia romana cronologicamente posposti alle vicende narrate nell'Eneide, mostrando doti profetiche.

Descrizione
Le immagini incise sullo scudo mostrano, attraverso la tecnica dell'ekphrasis, momenti della storia romana, da Romolo fino al trionfo di Augusto; tale descrizione occupa i versi 626-731 del libro VIII dell'Eneide. Lo scudo riporta la celebre immagine della lupa con i gemelli, il ratto delle Sabine e il supplizio di Mezio Fufezio; seguono la cacciata di Tarquinio il Superbo, lo scontro con Porsenna, rapidi accenni agli episodi di Orazio Coclite dell'eroina Clelia, della Rupe Tarpea e dei Galli disturbati dalle oche del Campidoglio. Le matrone romane assistono al corteo sacro dei sacerdoti Luperci, Salii e Flamini. Segue un importante salto temporale, con la visione del Tartaro con Catilina spaventato dalle Furie e i Campi Elisi con Catone che legifera tra le anime giuste.
Al centro, tra una cornice di delfini d'argento che guizzano in un mare dorato, è raffigurata la battaglia navale di Azio, che vede scontrarsi non solo Augusto e Agrippa contro Antonio e Cleopatra, ma anche gli dei greci contro gli dei egizi, con Anubis e il Nilo che accoglie i vinti. Ultima scena è il trionfo finale di Ottaviano Augusto a Roma, che assiste in trono al  corteo dei popoli sottomessi e dedica le spoglie dei vinti nei templi.

Rapporto con Omero
La descrizione dello scudo di Enea è chiaramente mutuata da quella dello Scudo di Achille, contenuta nell'Iliade di Omero; Virgilio si discosta dall'esempio greco per il soggetto di cui si occupa, una visione futura (ovviamente a posteriori) dell'intera storia di Roma, legando i trionfi del contemporaneo imperatore Augusto al mito dell'eroe Enea, di cui è effigiata "tutta la discendenza futura dal sangue di Ascanio".

Note

Voci correlate
Scudo di Achille

Eneide
Mitologia romana


text:
Luigi Taparelli D'Azeglio (Turin, Italia, 24 de noviembre de 1793 - Roma, 21 de septiembre de 1862) fue un sacerdote italiano de la Compañía de Jesús. Filósofo y teólogo, fue quien acuñó el término justicia social.

Fe y conciencia social

Estudió primero derecho, se ordenó sacerdote jesuita en 1820 tras haber hecho el noviciado en San Andrés del Quirinal, en Roma. Su pensamiento, de contenido social, considera que la autoridad emana de Dios y del pueblo, pero la sociedad no establece la jerarquía, sino la necesidad de los ciudadanos. Para él, la sociedad se compone del poder público y la obligación de cooperar con el bien social. Llega a la hipótesis de una sociedad establecida bajo un orden social generado por la aptitud y obligación del hombre para la vida en comunidad. Ejerció también como profesor e investigador del derecho natural, su pensamiento se anticipa a la concepción moderna de una comunidad internacional.

Fundó junto a Carlo Maria Curci la revista La Civiltà Cattolica. Se considera que su obra influyó en el papa Leon XIII quien publicó la Encíclica Rerum Novarum.

Publicaciones
 Ensayo teórico del derecho natural fundado sobre los hechos (1840-1843)
 Examen crítico de los ordenamientos representativos en la sociedad moderna (1854).
 Las razones de lo bello según los principios de santo Tomás de Aquino (1860)
 Curso elemental sobre el derecho natural al uso de las escuelas.

Véase también
 Rerum Novarum

Enlaces externos
 La Civiltà Cattolica

Religiosos de Italia del siglo XIX

Sacerdotes católicos de Italia del siglo XIX
Jesuitas de Italia
Nacidos en Turín
Fallecidos en Roma


In this, the second in our series of guides to the 2015 Heritage Open Days, we take a look at a few of the highlights that can be found in Cheshire, a county that is famous for its many white plaster and black timber-frame houses.
Cheshire's heritage riches also include some of Britain's grandest houses including Arley Hall, Dunham Massey Hall and Tatton Park. And if that weren't enough, the County town, Chester, has a total of 550 listed buildings. Our Cheshire highlights include a silk mill that inspired the grand-daddy of modern day computing, the finest timber frame building in Chester and a brewing magnate's Jacobean residence.
During the Heritage Open Days, there will be demonstrations featuring hi-tech 3D modelling of Roman Chester, with visualisations of the City's Roman fort based on the latest archaeological findings.
No 8 White Friars is a Jacobean private residence in Chester with Grade II listing. It was built in the 16th century and many of the original Jacobean features can be explored and viewed. It also features a heraldic over mantel which displays the arms of the Worshipful Company of Brewers, which ties in with the evidence which suggests the House was owned by a brewer by the name of Richard Harrison. During the 3 day Heritage Open Days event, the House opens up for four tours on both 12 and 13 September.
It may not look like much from the outside but within the walls of the Paradise Mills in Macclesfield is a truly stunningly preserved piece of history. The Mill is home to 26 original and fully working Jacquard looms. These looms, designed by Frenchman Joseph Marie Jacquard, are considered by many to be the forerunner of modern day computers, using as they did punch cards to produce automatically complex patterns. Amazingly Paradise Mill produced silk right up until its closure in 1981, when cheap imports and new materials forced its closure.
During the Heritage Open Days, the Mill's normal admission charges will be waived and there'll be regular tours led by an experienced guide, offering real insight into the way the mill used to run.
Even though we have only dipped our toe into what's on offer in Cheshire during Heritage Open Days, I hope that the above has whetted your appetite. There's a total of 99 Cheshire properties listed on the Event's website. You can find out more about all of them by going to the Heritage Open Days website, clicking on the 'Explore' tab and selecting Cheshire from the drop down menu.
We currently have a number of deals in Cheshire with availability (at the time of writing) over the Heritage Open Days Weekend (10 - 13 September 2015). Stay at the Cranage Hall which offers a two course evening meal, plus a ticket to explore the 1000 acres of Tatton Park. Or stay at The Brook Meadow Hotel which includes a 3 course meal, as well as 10% off Cheshire Oaks Designer Outlet.