Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\deg f>\deg g$. Suppose that for infinitely many primes $p$ the polynomial $pf+g$ has a rational root. Prove that $f$ has a rational root.

Let $P$ be a cubic polynomial with $P(0) = k, P(1) = 2k,$ and $P(-1) = 3k$. What is $P(2) + P(-2)$? $ \textbf{(A) }0 \qquad\textbf{(B) }k \qquad\textbf{(C) }6k \qquad\textbf{(D) }7k\qquad\textbf{(E) }14k\qquad $

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? \[ \begin{picture}(60,60) \put(5,3){\put(3,0){\line(6,0){8}} \put(17,0){\line(6,0){8}} \put(31,0){\line(6,0){8}} \put(45,0){\line(6,0){8}} \put(10,14){\line(6,0){8}} \put(24,14){\line(6,0){8}} \put(38,14){\line(6,0){8}} \put(17,28){\line(6,0){8}} \put(31,28){\line(6,0){8}} \put(24,42){\line(6,0){8}} \put(1,2){\line(1,2){5}} \put(15,2){\line(1,2){5}} \put(29,2){\line(1,2){5}} \put(43,2){\line(1,2){5}} \put(8,16){\line(1,2){5}} \put(22,16){\line(1,2){5}} \put(36,16){\line(1,2){5}} \put(15,30){\line(1,2){5}} \put(29,30){\line(1,2){5}} \put(22,44){\line(1,2){5}} \put(13,2){\line( \minus{} 1,2){5}} \put(27,2){\line( \minus{} 1,2){5}} \put(41,2){\line( \minus{} 1,2){5}} \put(55,2){\line( \minus{} 1,2){5}} \put(20,16){\line( \minus{} 1,2){5}} \put(34,16){\line( \minus{} 1,2){5}} \put(48,16){\line( \minus{} 1,2){5}} \put(27,30){\line( \minus{} 1,2){5}} \put(41,30){\line( \minus{} 1,2){5}} \put(34,44){\line( \minus{} 1,2){5}} \put(0,0){\circle{6}} \put(14,0){\circle{6}} \put(28,0){\circle{6}} \put(42,0){\circle{6}} \put(56,0){\circle{6}} \put(7,14){\circle{6}} \put(21,14){\circle{6}} \put(35,14){\circle{6}} \put(49,14){\circle{6}} \put(14,28){\circle{6}} \put(28,28){\circle{6}} \put(42,28){\circle{6}} \put(21,42){\circle{6}} \put(35,42){\circle{6}} \put(28,56){\circle{6}}} \end{picture}\]

Determine $w^2+x^2+y^2+z^2$ if \[ \begin{array}{l} \displaystyle \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 \\ \displaystyle \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 \\ \displaystyle \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 \\ \displaystyle \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 \\ \end{array} \]

Let the roots $a,b,c$ of $$f(x)=x^3 +p x^2 + qx+r$$ be real, and let $a\leq b\leq c$. Prove that $f'(x)$ has a root in the interval $\left[\frac{b+c}{2}, \frac{b+2c}{3}\right]$. What will be the form of $f(x)$ if the root in question falls at either end of the interval?

Let $r$ and $s$ be positive integers. Define $a_0 = 0$, $a_1 = 1$, and $a_n = ra_{n-1} + sa_{n-2}$ for $n \geq 2$. Let $f_n = a_1a_2\cdots a_n$. Prove that $\displaystyle\frac{f_n}{f_kf_{n-k}}$ is an integer for all integers $n$ and $k$ such that $0 < k < n$. [i]Evan O' Dorney.[/i]

Consider a polynomial $P(x) = \prod^9_{j=1}(x+d_j),$ where $d_1, d_2, \ldots d_9$ are nine distinct integers. Prove that there exists an integer $N,$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20. [i]Proposed by Luxembourg[/i]

The two polynomials $(x) =x^4+ax^3+bx^2+cx+d$ and $Q(x) = x^2+px+q$ take negative values on an interval $I$ of length greater than $2$, and nonnegative values outside of $I$. Prove that there exists $x_0 \in \mathbb R$ such that $P(x_0) < Q(x_0)$.

The polynomial $P(x)=3x^3-2x^2+ax-b$ has roots $\sin^2\theta$, $\cos^2\theta$, and $\sin\theta\cos\theta$ for some angle $\theta$. Find $P(1)$.

Let $P$ be a quadratic (polynomial of degree two) with a positive leading coefficient and negative discriminant. Prove that there exists three quadratics $P_1, P_2, P_3$ such that: - $P(x) = P_1(x) + P_2(x) + P_3(x)$ - $P_1, P_2, P_3$ have positive leading coefficients and zero discriminants (and hence each has a double root) - The roots of $P_1, P_2, P_3$ are different

Let $n$ be a positive integer and $P(x)$ be a polynomial with integer coefficients such that $P(1)=1,P(2)=2,\ldots,P(n)=n$. Prove that $P(0)$ is divisible by $2 \cdot 3 \cdot \ldots \cdot n$. [i]A. Voidelevich[/i]

Let $P(x)$ be a monic polynomial of degree $100$ with $100$ distinct noninteger real roots. Suppose that each of polynomials $P(2x^2 - 4x)$ and $P(4x - 2x^2)$ has exactly $130$ distinct real roots. Prove that there exist non constant polynomials $A(x),B(x)$ such that $A(x)B(x) = P(x)$ and $A(x) = B(x)$ has no root in $(-1.1)$

Suppose that $x$ and $y$ are complex numbers such that \[\frac{x^{n}-y^{n}}{x-y}\] are integers for some four consecutive positive integers $n$. Prove that it is an integer for all positive integers $n$.

Find $x^2+y^2$ if $x$ and $y$ are positive integers such that \[xy+x+y = 71\qquad\text{and}\qquad x^2y+xy^2 = 880.\]

$p(x)$ is a polynomial such that $p(y^2+1) = 6y^4 - y^2 + 5$. Find $p(y^2-1)$.

Prove that if $ c, d $ are integers with $ c \neq d $, $ d > 0 $ then the equation $$ x^3 - 3cx^2 - dx + c = 0$$ has no more than one rational root.

Let $f$ be a primitive polynomial with integral coefficients (their highest common factor is $1$ ) such that $f$ is irreducible in $\mathbb{Q}[X]$ , and $f(X^2)$ is reducible in $\mathbb{Q}[X]$ . Show that $f= \pm(u^2-Xv^2)$ for some polynomials $u$ and $v$ with integral coefficients.

Let $\mathcal K$ be a field with $2^{n}$ elements, $n \in \mathbb N^\ast$, and $f$ be the polynomial $X^{4}+X+1$. Prove that: (a) if $n$ is even, then $f$ is reducible in $\mathcal K[X]$; (b) if $n$ is odd, then $f$ is irreducible in $\mathcal K[X]$. [hide="Remark."]I saw the official solution and it wasn't that difficult, but I just couldn't solve this bloody problem.[/hide]

If $q_1 ( x )$ and $r_ 1$ are the quotient and remainder, respectively, when the polynomial $x^ 8$ is divided by $x + \tfrac{1}{2}$ , and if $q_ 2 ( x )$ and $r_2$ are the quotient and remainder, respectively, when $q_ 1 ( x )$ is divided by $x + \tfrac{1}{2}$, then $r_2$ equals $\textbf{(A) }\frac{1}{256}\qquad\textbf{(B) }-\frac{1}{16}\qquad\textbf{(C) }1\qquad\textbf{(D) }-16\qquad\textbf{(E) }256$

Let $ p(x)$ be a cubic polynomial with rational coefficients. $ q_1$, $ q_2$, $ q_3$, ... is a sequence of rationals such that $ q_n \equal{} p(q_{n \plus{} 1})$ for all positive $ n$. Show that for some $ k$, we have $ q_{n \plus{} k} \equal{} q_n$ for all positive $ n$.

Let $P(x)$ be a polynomial with integer coefficients. Show that if $Q(x) = P(x) +12$ has at least six distinct integer roots, then $P(x)$ has no integer roots.

Let $P(x)$ be a polynomial taking integer values at integer inputs. Are there infinitely many natural numbers that are not representable in the form $P(k)-2^n$ where $n{}$ and $k{}$ are non-negative integers? [i]Proposed by F. Petrov[/i]

Given a polynomial $P(x)=a_{d}x^{d}+ \ldots +a_{2}x^{2}+a_{0}$ with positive integers for coefficients and degree $d\geq 2$. Consider the sequence defined by $$b_{1}=a_{0} ,b_{n+1}=P(b_{n}) $$ for $n \geq 1$ . Prove that for all $n \geq 2$ there exists a prime $p$ such that $p$ divides $b_{n}$ but does not divide $b_{1}b_{2} \ldots b_{n-1}$.

We have $n$ points in the plane, no three on a line. We call $k$ of them good if they form a convex polygon and there is no other point in the convex polygon. Suppose that for a fixed $k$ the number of $k$ good points is $c_k$. Show that the following sum is independent of the structure of points and only depends on $n$ : \[ \sum_{i=3}^n (-1)^i c_i \]

Is there a triangle with angles in ratio of $ 1: 2: 4$ and the length of its sides are integers with at least one of them is a prime number? [i]Nanang Susyanto, Jogjakarta[/i]