A polynomial $x^n + a_1x^{n-1} + a_2x^{n-2} +... + a_{n-2}x^2 + a_{n-1}x + a_n$ has $n$ distinct real roots $x_1, x_2,...,x_n$, where $n > 1$. The polynomial $nx^{n-1}+ (n - 1)a_1x^{n-2} + (n - 2)a_2x^{n-3} + ...+ 2a_{n-2}x + a_{n-1}$ has roots $y_1, y_2,..., y_{n_1}$. Prove that $\frac{x^2_1+ x^2_2+ ...+ x^2_n}{n}>\frac{y^2_1 + y^2_2 + ...+ y^2_{n-1}}{n - 1}$

For any positive integer $n$ and $0 \leqslant i \leqslant n$, denote $C_n^i \equiv c(n,i)\pmod{2}$, where $c(n,i) \in \left\{ {0,1} \right\}$. Define \[f(n,q) = \sum\limits_{i = 0}^n {c(n,i){q^i}}\] where $m,n,q$ are positive integers and $q + 1 \ne {2^\alpha }$ for any $\alpha \in \mathbb N$. Prove that if $f(m,q)\left| {f(n,q)} \right.$, then $f(m,r)\left| {f(n,r)} \right.$ for any positive integer $r$.

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

Prove $ \ \ \ \frac{22}{7}\minus{}\pi \equal{}\int_0^1 \frac{x^4(1\minus{}x)^4}{1\plus{}x^2}\ dx$.

Does there exist positive reals $a_0, a_1,\ldots ,a_{19}$, such that the polynomial $P(x)=x^{20}+a_{19}x^{19}+\ldots +a_1x+a_0$ does not have any real roots, yet all polynomials formed from swapping any two coefficients $a_i,a_j$ has at least one real root?

When the roots of the polynomial \[P(x)=\prod_{i=1}^{10}(x-i)^{i}\] are removed from the real number line, what remains is the union of $11$ disjoint open intervals. On how many of those intervals is $P(x)$ positive? $\textbf{(A)}~3\qquad\textbf{(B)}~4\qquad\textbf{(C)}~5\qquad\textbf{(D)}~6\qquad\textbf{(E)}~7$

Given an expression $x^2 + ax + b$ where $a,b$ are integer coefficients. At any step, one can change the expression by adding either $1$ or $-1$ to only one of the two coefficients $a, b$. a) Suppose that the initial expression has $a =-7$ and $b = 19$. Show your modification steps to obtain a new expression that has zero value at some integer value of $x$. b) Starting from the initial expression as above, one gets the expression $x^2 - 17x + 9$ after $m$ modification steps. Prove that at a certain step $k$ with $k < m$, the obtained expression has zero value at some integer value of $x$.

Let $n \geq 2$ be a fixed positive integer and let $a_{0},a_{1},...,a_{n-1}$ be real numbers. Assume that all of the roots of the polynomial $P(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{1}x+a_{0}$ are strictly positive real numbers. Determine the smallest possible value of $\frac{a_{n-1}^{2}}{a_{n-2}}$ over all such polynomials. [i]Proposed by Nikola Velov[/i]

Show that if the integers $a_1$; $\dots$ $a_m$ are nonzero and for each $k =0; 1; \dots ;n$ ($n < m - 1$), $a_1 + a_22^k + a_33^k + \dots + a_mm^k = 0$; then the sequence $a_1, \dots, a_m$ contains at least $n+1$ pairs of consecutive terms having opposite signs. [i]O. Musin[/i]

Let $f(x)=x^{n}+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Let $p(x)$ be a polynomial of degree 4 with leading coefficients 1. Suppose $p(1)=1$, $p(2)=2$, $p(3)=3$, $p(4)=4$. Then $p(5)=$ [list=1] [*] 5 [*] $\frac{25}{6}$ [*] 29 [*] 35 [/list]

Let $P(x)$ = $a_0+a_1x+a_2x^2+\cdots +a_nx^n$ be a polynomial in which $a_i$ is non-negative integer for each $i \in$ {$0,1,2,3,....,n$} . If $P(1) = 4$ and $P(5) = 136$, what is the value of $P(3)$?

We call a monic polynomial $P(x) \in \mathbb{Z}[x]$ [i]square-free mod n[/i] if there [u]dose not[/u] exist polynomials $Q(x),R(x) \in \mathbb{Z}[x]$ with $Q$ being non-constant and $P(x) \equiv Q(x)^2 R(x) \mod n$. Given a prime $p$ and integer $m \geq 2$. Find the number of monic [i]square-free mod p[/i] $P(x)$ with degree $m$ and coeeficients in $\{0,1,2,3,...,p-1\}$. [i]Proposed by Masud Shafaie[/i]

Suppose $f(z)=z^n+a_1z^{n-1}+...+a_n$ for which $a_1,a_2,...,a_n\in \mathbb C$. Prove that the following polynomial has only one positive real root like $\alpha$ \[x^n+\Re(a_1)x^{n-1}-|a_2|x^{n-2}-...-|a_n|\] and the following polynomial has only one positive real root like $\beta$ \[x^n-\Re(a_1)x^{n-1}-|a_2|x^{n-2}-...-|a_n|.\] And roots of the polynomial $f(z)$ satisfy $-\beta \le \Re(z) \le \alpha$.

Do there exist polynomials $a(x)$, $b(x)$, $c(y)$, $d(y)$ such that \[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\] holds identically?