The functions $ f(x) ,\ g(x)$ satify that $ f(x) \equal{} \frac {x^3}{2} \plus{} 1 \minus{} x\int_0^x g(t)\ dt,\ g(x) \equal{} x \minus{} \int_0^1 f(t)\ dt$. Let $ l_1,\ l_2$ be the tangent lines of the curve $ y \equal{} f(x)$, which pass through the point $ (a,\ g(a))$ on the curve $ y \equal{} g(x)$. Find the minimum area of the figure bounded by the tangent tlines $ l_1,\ l_2$ and the curve $ y \equal{} f(x)$ .

The sequence of positive integers $a_1$, $a_2$,$...$ is such that for each $n = 1$,$2$, $...$ the quadratic equation $$a_{n+2}x^2 + a_{n+1}x+ a_n = 0$$ has a real root. Can the sequence consist of (a) $10 $ terms, (b) an infinite number of terms? (A Shapovalov)

For any polynomial $P(x)=a_0+a_1x+\ldots+a_kx^k$ with integer coefficients, the number of odd coefficients is denoted by $o(P)$. For $i-0,1,2,\ldots$ let $Q_i(x)=(1+x)^i$. Prove that if $i_1,i_2,\ldots,i_n$ are integers satisfying $0\le i_1<i_2<\ldots<i_n$, then: \[ o(Q_{i_{1}}+Q_{i_{2}}+\ldots+Q_{i_{n}})\ge o(Q_{i_{1}}). \]

Find the greatest $n$ such that $(z+1)^n = z^n + 1$ has all its non-zero roots in the unitary circumference, e.g. $(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0$ implies $|\alpha| = 1.$

Let $ a_1,a_{2},\ldots ,a_{100}$ be nonnegative integers satisfying the inequality \[a_1\cdot (a_1-1)\cdot\ldots\cdot (a_1-20)+a_2\cdot (a_2-1)\cdot\ldots\cdot (a_2-20)+\\ \ldots+a_{100}\cdot (a_{100}-1)\cdot\ldots\cdot (a_{100}-20)\le 100\cdot 99\cdot 98\cdot\ldots\cdot 79.\] Prove that $a_1+a_2+\ldots+a_{100}\le 9900$.

We say that a ring $A$ has property $(P)$ if any non-zero element can be written uniquely as the sum of an invertible element and a non-invertible element. a) If in $A$, $1+1=0$, prove that $A$ has property $(P)$ if and only if $A$ is a field. b) Give an example of a ring that is not a field, containing at least two elements, and having property $(P)$. [i]Dan Schwarz[/i]

We define the [i]Fibonacci sequence[/i] $\{F_n\}_{n\ge0}$ by $F_0=0$, $F_1=1$, and for $n\ge2$, $F_n=F_{n-1}+F_{n-2}$; we define the [i]Stirling number of the second kind[/i] $S(n,k)$ as the number of ways to partition a set of $n\ge1$ distinguishable elements into $k\ge1$ indistinguishable nonempty subsets. For every positive integer $n$, let $t_n = \sum_{k=1}^{n} S(n,k) F_k$. Let $p\ge7$ be a prime. Prove that \[ t_{n+p^{2p}-1} \equiv t_n \pmod{p} \] for all $n\ge1$. [i]Proposed by Victor Wang[/i]

The game involves two players $A$ and $B$. Player $A$ sets the value of one of the coefficients $a, b$ or $c$ of the polynomial $$x^3 + ax^2 + bx + c.$$ Player $B$ indicates the value of any of the two remaining coefficients . Player $A$ then sets the value of the last coefficients. Is there a strategy for player A such that no matter how player $B$ plays, the equation $$x^3 + ax^2 + bx + c = 0$$ to have three different (real) solutions?

Prove that for any quadratic polynomial $f(x)=x^2+px+q$ with integer coefficients, it is possible to find another polynomial $q(x)=2x^2+rx+s$ with integer coefficients so that \[\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset .\]

Prove that $$x^4 + 2x^2 + 2x + 2$$ is not the product of two polynomials $x^2 + ax + b$ and $x^2 + cx + d$ in which $a$, $b$, $c$, $d$ are integers.

Let $q_{0},q_{1},...$ be a sequence of integers such that a) for any $m>n$ we have $m-n\mid q_{m}-q_{n}$, and b) $|q_{n}|\leq n^{10}, \ \forall n\geq 0$. Prove there exists a polynomial $Q$ such that $q_{n}=Q(n), \ \forall n\geq 0$.

Consider the locus given by the real polynomial equation $$ Ax^2 +Bxy+Cy^2 +Dx^3 +E x^2 y +F xy^2 +G y^3=0,$$ where $B^2 -4AC <0.$ Prove that there is a positive number $\delta$ such that there are no points of the locus in the punctured disk $$0 <x^2 +y^2 < \delta^2.$$

For which positive integers $n$ do there exist two polynomials $f$ and $g$ with integer coefficients of $n$ variables $x_1, x_2, \ldots , x_n$ such that the following equality is satisfied: \[\sum_{i=1}^n x_i f(x_1, x_2, \ldots , x_n) = g(x_1^2, x_2^2, \ldots , x_n^2) \ ? \]

One considers for $n > 2$ the polynomial: $$(x^2-x+1)^n - (x^2-x+2)^n+ (1+x)^n+(2-x)^n$$ Show that the degree of this polynomial is $2n - 2$. The polynomial is written in the form $$a_0+a_1x+a_2x^2+...+a_{2n-2}x^{2n-2}$$ Prove that $a_2+a_3+...+a_{2n-2}=0$

$k$ and $n$ are positive integers that are greater than $1$. $N$ is the set of positive integers. $A_1, A_2, \cdots A_k$ are pairwise not-intersecting subsets of $N$ and $A_1 \cup A_2 \cup \cdots \cup A_k = N$. Prove that for some $i \in \{ 1,2,\cdots,k \}$, there exsits infinity many non-factorable n-th degree polynomials so that coefficients of one polynomial are pairwise distinct and all the coeficients are in $A_i$.

Find all pairs of real number $x$ and $y$ which simultaneously satisfy the following 2 relations: $x+y+4=\frac{12x+11y}{x^2+y^2}$ $y-x+3=\frac{11x-12y}{x^2+y^2}$

Find all triples $(a,b,c)$ of real numbers such that the following system holds: $$\begin{cases} a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{cases}$$ [i]Proposed by Dorlir Ahmeti, Albania[/i]

Let $A$ be an $n\times n$ matrix of real numbers for some $n\ge 1.$ For each positive integer $k,$ let $A^{[k]}$ be the matrix obtained by raising each entry to the $k$th power. Show that if $A^k=A^{[k]}$ for $k=1,2,\cdots,n+1,$ then $A^k=A^{[k]}$ for all $k\ge 1.$

Let $\mathbb{R}[x,y]$ denote the set of two-variable polynomials with real coefficients. We say that the pair $(a,b)$ is a [i]zero[/i] of the polynomial $f \in \mathbb{R}[x,y]$ if $f(a,b)=0$. If polynomials $p,q \in \mathbb{R}[x,y]$ have infinitely many common zeros, does it follow that there exists a non-constant polynomial $r \in \mathbb{R}[x,y]$ which is a factor of both $p$ and $q$?

Let $P$ be a polynomial with integer coefficients. There exist integers $a$ and $b$ such that $P(a) \cdot P(b)=-(a-b)^2$. Prove that $P(a)+P(b)=0$.

Two players by turn paint the circles on the given picture each with his colour. At the end, the rest of the area of each of small triangles is painted by the colour of the majority of vertices of this triangle. The winner is one who gets larger area of his colour (the area of circles is taken into account). Does any of them have winning strategy? If yes, then who wins? \[ \begin{picture}(60,60) \put(5,3){\put(3,0){\line(6,0){8}} \put(17,0){\line(6,0){8}} \put(31,0){\line(6,0){8}} \put(45,0){\line(6,0){8}} \put(10,14){\line(6,0){8}} \put(24,14){\line(6,0){8}} \put(38,14){\line(6,0){8}} \put(17,28){\line(6,0){8}} \put(31,28){\line(6,0){8}} \put(24,42){\line(6,0){8}} \put(1,2){\line(1,2){5}} \put(15,2){\line(1,2){5}} \put(29,2){\line(1,2){5}} \put(43,2){\line(1,2){5}} \put(8,16){\line(1,2){5}} \put(22,16){\line(1,2){5}} \put(36,16){\line(1,2){5}} \put(15,30){\line(1,2){5}} \put(29,30){\line(1,2){5}} \put(22,44){\line(1,2){5}} \put(13,2){\line( \minus{} 1,2){5}} \put(27,2){\line( \minus{} 1,2){5}} \put(41,2){\line( \minus{} 1,2){5}} \put(55,2){\line( \minus{} 1,2){5}} \put(20,16){\line( \minus{} 1,2){5}} \put(34,16){\line( \minus{} 1,2){5}} \put(48,16){\line( \minus{} 1,2){5}} \put(27,30){\line( \minus{} 1,2){5}} \put(41,30){\line( \minus{} 1,2){5}} \put(34,44){\line( \minus{} 1,2){5}} \put(0,0){\circle{6}} \put(14,0){\circle{6}} \put(28,0){\circle{6}} \put(42,0){\circle{6}} \put(56,0){\circle{6}} \put(7,14){\circle{6}} \put(21,14){\circle{6}} \put(35,14){\circle{6}} \put(49,14){\circle{6}} \put(14,28){\circle{6}} \put(28,28){\circle{6}} \put(42,28){\circle{6}} \put(21,42){\circle{6}} \put(35,42){\circle{6}} \put(28,56){\circle{6}}} \end{picture}\]

Let $P(x, y)$ be a non-constant homogeneous polynomial with real coefficients such that $P(\sin t, \cos t) = 1$ for every real number $t$. Prove that there exists a positive integer $k$ such that $P(x, y) = (x^2 + y^2)^k$.

Given the equation $x^2+ax+1=0$, determine: a) The interval of possible values for $a$ where the solutions to the previous equation are not real. b) The loci of the roots of the polynomial, when $a$ is in the previous interval.

Determine the pairs of sets $X,Y\subset\mathbb{R}$ for which the following is true: if $f(x, y)$ is a function on $X\times Y{}$ such that for every $x\in X$ it is equal to a polynomial in $y$ on $Y$ and for every $y\in Y$ it is equal to a polynomial in $x$ on $X$ then $f$ is a bivariate polynomial on $X\times Y.$

Let $Q(x)$ be a polynomial with integer coefficients. Prove that there exists a polynomial $P(x)$ with integer coefficients such that for every integer $n\ge\deg{Q}$, \[\sum_{i=0}^{n}\frac{!i P(i)}{i!(n-i)!} = Q(n),\]where $!i$ denotes the number of derangements (permutations with no fixed points) of $1,2,\ldots,i$. [i]Calvin Deng.[/i]