Školjka

%V0 $(FRA 1)$ Let $a$ and $b$ be two nonnegative integers. Denote by $H(a, b)$ the set of numbers $n$ of the form $n = pa + qb,$ where $p$ and $q$ are positive integers. Determine $H(a) = H(a, a)$. Prove that if $a \neq b,$ it is enough to know all the sets $H(a, b)$ for coprime numbers $a, b$ in order to know all the sets $H(a, b)$. Prove that in the case of coprime numbers $a$ and $b, H(a, b)$ contains all numbers greater than or equal to $\omega = (a - 1)(b -1)$ and also $\frac{\omega}{2}$ numbers smaller than $\omega$

%V0 $(FRA 6)$ Consider the integer $d = \frac{a^b-1}{c}$, where $a, b$, and $c$ are positive integers and $c \le a.$ Prove that the set $G$ of integers that are between $1$ and $d$ and relatively prime to $d$ (the number of such integers is denoted by $\phi(d)$) can be partitioned into $n$ subsets, each of which consists of $b$ elements. What can be said about the rational number $\frac{\phi(d)}{b}?$

%V0 $(GBR 1)$ The polynomial $P(x) = a_0x^k + a_1x^{k-1} + \cdots + a_k$, where $a_0,\cdots, a_k$ are integers, is said to be divisible by an integer $m$ if $P(x)$ is a multiple of $m$ for every integral value of $x$. Show that if $P(x)$ is divisible by $m$, then $a_0 \cdot k!$ is a multiple of $m$. Also prove that if $a, k,m$ are positive integers such that $ak!$ is a multiple of $m$, then a polynomial $P(x)$ with leading term $ax^k$can be found that is divisible by $m.$

%V0 $(GBR 2)$ Let $a, b, x, y$ be positive integers such that $a$ and $b$ have no common divisor greater than $1$. Prove that the largest number not expressible in the form $ax + by$ is $ab - a - b$. If $N(k)$ is the largest number not expressible in the form $ax + by$ in only $k$ ways, find $N(k).$

%V0 Let $S = \{1,2,3,\cdots ,280\}$. Find the smallest integer $n$ such that each $n$-element subset of $S$ contains five numbers which are pairwise relatively prime.

%V0 Let $a, b, c$ be integers and $p$ an odd prime number. Prove that if $f(x) = ax^2 + bx + c$ is a perfect square for $2p - 1$ consecutive integer values of $x,$ then $p$ divides $b^2 - 4ac.$