\documentclass{article}
\usepackage{amssymb, amsmath, amsthm}
\theoremstyle{definition}
\newtheorem{zadA}{$\blacklozenge$ A}
\newtheorem{zadB}{$\blacklozenge$ B}
\DeclareMathOperator{\amod}{amod}
\newcommand*{\hm}[1]{#1\nobreak\discretionary{}%
            {\hbox{$\mathsurround=0pt #1$}}{}}

\begin{document}
\centerline{\bfseries \Large Hints and solutions to the problems
from sections A and B}
\begin{zadA}
a) $a-2x=0$
\\
b) $a-2x-1=0$
\\
c) $a-x^2=0$
\\
d) $a-x^3=0$
\end{zadA}

\begin{zadA}
$
 \left\{
  \begin{aligned}
   \mathfrak D_1(x_1, \ldots , x_n)&=0\\
   \ldots \\
   \mathfrak D_m(x_1, \ldots , x_n)&=0\\
  \end{aligned}
 \right.
$ $\Longleftrightarrow$ $\mathfrak D_1(x_1, \ldots , x_n)^2+\ldots
+\mathfrak D_m(x_1, \ldots , x_n)^2=0$
\end{zadA}

\begin{zadA}
If the sets $\mathbb A\hm=\{ \mathfrak F(x_1, \ldots , x_n)=0\}$
and $\mathbb B=\{ \mathfrak G(x_1, \ldots , x_n)=0\}$ then
$\mathbb A\cap \mathbb B\hm=\{\mathfrak F(x_1, \ldots ,
x_n)\cdot\mathfrak G(x_1, \ldots , x_n)=0$ and $\mathbb
A\cup\mathbb B\hm={\mathfrak F(x_1, \ldots , x_n)^2+\mathfrak
G(x_1, \ldots , x_n)^2=0}$.
\end{zadA}

\begin{zadA}

\end{zadA}

\begin{zadA}
a) $a=b+x+1$
\\
b) $a=b\cdot x$
\\
c) $a=c\cdot x+b$ and $b<c$
\\
d)
$
 \left[
  \begin{aligned}
   \left\{
    \begin{aligned}
     a&=c\cdot x+b\\
     b&\leqslant c-b\\
    \end{aligned}
   \right.\\
   \left\{
    \begin{aligned}
     a&=c\cdot x-b\\
     b&<c-b\\
    \end{aligned}
   \right.\\
  \end{aligned}
 \right.
$
\\
\\
e) $b=c\cdot a+x$ and $x<c$
\end{zadA}

\begin{zadA}
$a(x_1-x_2)+b(y_1-y_2)=1$ if and only if then $a$ and $b$
co-prime.
\\
\\
$a=kt$, $b=lt$, $GCD(k, l)=1$ if and only if then $t=GCD(a, b)$.
\\
\\
$LCM(a, b)=\frac{ab}{GCD(a, b)}$
\end{zadA}

\begin{zadA}
a) $x^2<a<(x+1)^2$
\end{zadA}

\begin{zadA}
a) $d=k^2 \Rightarrow (x-ky)(x+ky)=1 \Rightarrow x-dy=x+dy
\Rightarrow y=0 \Rightarrow x=\pm 1$
\\
\\
b) Since $u_3-v_3\sqrt{d}=(u_1-v_1\sqrt{d})\cdot
(u_2-v_2\sqrt{d})$, then
$u_3^2-v_3^2d=(u_1^2-v_1^2d)(u_2^2-v_2^2d)=1$
\\
\\
c) $|x'|=\sqrt{1+y'^2d} \Rightarrow |x'|+|y'|=\sqrt{1+y'^2d}+|y'|$
is monotonic by $|y'|$. Let
$(x+y\sqrt{d})^n<x'+y'\sqrt{d}<(x+y\sqrt{d})^{n+1}$ ($x>0,\ y>0,\
x'>0,\ y'>0$). Multiplying it by $(x-y\sqrt{d})^n>0$ we get
$1<a+b\sqrt{d}<x+y\sqrt{d}$, such that $a^2-b^2d=1$. Since
$0<a+b\sqrt{d}$, then $a-b\sqrt{d}>0$. Since
$a-b\sqrt{d}<1<a+b\sqrt{d}$, then $b>0$. Since $a-b\sqrt{d}>0$,
then $a>0$. So, since while $|x|+|y|$ is increasing, $|y|$ is
increasing, $|x|+|y|\sqrt{d}$ is increasing, then $(x, y)$ is not
a minimal solution.
\end{zadA}

\begin{zadA}
a) See the previous solution.
\\
\\
b) This statement (and that is $x_n\equiv 1\pmod{k-1}$) can be
proved by induction on $n$.
\\
\\
c) It can be solved as the particular case of the next problem:
$1=x^2\hm-(\frac{b^2}4\hm-1)y^2\hm=(x+\frac b2y)^2-b(x+\frac
b2y)y+y^2$
\end{zadA}

\begin{zadA}
Consider the minimal solution and prove that every other solution
can be constructed as the iteration of
\end{zadA}

\begin{zadA}
This can be proved by induction of $n$.
\end{zadA}

\begin{zadA}
This can be proved by induction of $l$.
\end{zadA}

\begin{zadA}
This can be proved by using the previous problem (since
$\alpha_{km}(b)$ divides on $\alpha_{m}(b)$,
$\alpha_{4m-1}(b)=\alpha_{2m-1}(b)(\alpha_{2m}(b)-\alpha_{2m-2}(b))$
and
$\alpha_{2m}(b)=\alpha_m(b)(\alpha_{m+1}(b)-\alpha_{m-1}(b))$).
\end{zadA}

\begin{zadA}
This can be proved by induction on $n$.
\end{zadA}

\begin{zadA}
No comments
\end{zadA}

\begin{zadA}
It is obvious that $\alpha_n(b)(\amod v)=\alpha_n(w)(\amod v)$ and
$n(\amod u)=\alpha_n(w)(\amod u)$. Therefore, since
$v>2\alpha_k(b)>2\alpha_n(b)$ and $u>2k>2n$, then we get the
statement we need
\end{zadA}

\begin{zadA}
No comments
\end{zadA}

\begin{zadA}
The inequality $k\leqslant\alpha_k(b)$ can be proved by induction
on $k$
\end{zadA}

\begin{zadA}
No comments
\end{zadA}

\begin{zadA}
This can be proved by induction on $k$
\end{zadA}

\begin{zadA}
This can be proved by induction on $n$
\end{zadA}

\begin{zadA}
$\frac{(bn+4)^c}{(n-1)^c}\geqslant
\frac{\alpha_{c+1}(bn+4)}{\alpha_{c+1}(n)}\geqslant
\frac{(bn+3)^c}{n^c}$. The left and right parts of this inequality
tend to $b^c$.
\end{zadA}

\begin{zadA}
No comments
\end{zadA}

\begin{zadB}
$c=\left[\frac a{b^{k-1}}\right]\pmod{b^k}$
\end{zadB}

\begin{zadB}
a) $b=2^n$, $a=(2^n+1)^n$
\\
b)
\end{zadB}

\begin{zadB}
$p$ is prime if and only if then $GCD(p, (p-1)!)=1$
\end{zadB}

\begin{zadB}
$(x_0+1)(1-D(x_0, x_1, \ldots , x_m)^2)-1\geqslant0$ if and only
if then $1-D(x_0, x_1, \ldots , x_m)^2>0$, i.e. $D(x_0, x_1,
\ldots , x_m)=0$. So, $a=(x_0+1)(1-0)-1=x_0$.
\end{zadB}

\begin{zadB}
It follows from the two previous problems
\end{zadB}

\begin{zadB}
a) The first addendum in this sum equals to the number of the
factors divisible by $p$, the second --- by $p^2$ etc.
\\
\\
b) $\sum(\left[\frac{m+n}{p^k}\right]-\left[\frac
m{p^k}\right]-\left[\frac n{p^k}\right])$ is equal to the demanded
number.
\end{zadB}

\begin{zadB}
$\binom{x_np^{n-1}+\ldots +x_1p+x_0}{y_np^{n-1}+\ldots +y_1p+y_0}$
divides on $p$ if and only if then there exists $i$, such that
$x_i<y_i$.
\end{zadB}

\begin{zadB}
$a<e\cdot\frac{p^n-1}{p-1}$
\end{zadB}

\begin{zadB}
$(y_nb_1^{n-1}+\ldots +y_1b_1+y_0)-(y_nb_2^{n-1}+\ldots
+y_1b_2+y_0)\equiv 0\pmod{b_2-b_1}$ (since
$b_1^k-b_2^k=(b_1-b_2)(b_1^{k-1}+b_1^{k-2}b_2+\ldots +b_2^{k-1})$)
\end{zadB}

\begin{zadB}
$a_1=a_2\pmod{b_2-b_1}$ (since $a_1<b_1^n<b_2-b_1$)
\end{zadB}

%\begin{zadB}
%Let $p$ be a prime such that $b_1^n<p-b_1$, $b_2^n<p-b_2$ (where
%$n$ is natural such that $b_1^{n-1}<a_1<b_1^n$). Let $x$ be such a
%number that
%\end{zadB}

\end{document}
