Find all polynomials $P$ with real coefficients such that for all $x, y ,z \in R$, $$P(x)+P(y)+P(z)+P(x+y+z)=P(x+y)+P(y+z)+P(z+x)$$

a) Let $a\in \mathbb{R}$ and $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous function for which there exists an antiderivative $F \colon \mathbb{R} \to \mathbb{R} $, such that $F(x)+a\cdot f(x) \geq 0$, for any $x \in \mathbb{R}$, and$ \lim_{|x| \to \infty} \frac{F(x)}{e^{|\alpha \cdot x|}}=0$ holds for any $\alpha \in \mathbb{R}^*$. Prove that $F(x) \geq 0$ for all $x \in \mathbb{R}$. b) Let $n\geq 2$ be a positive integer, $g \in \mathbb{R}[X]$, $g = X^n + a_1X^{n-1}+ \dots + a_{n-1}X+a_n$ be a polynomial with all of its roots being real, and $f \colon \mathbb{R} \to \mathbb{R}$ a polynomial function such that $f(x)+a_1\cdot f'(x)+a_2\cdot f^{(2)}(x)+\dots+a_n\cdot f^{(n)}(x) \geq 0$ for any $x \in \mathbb{R}$. Prove that $f(x) \geq 0$ for all $x \in \mathbb{R}$.

There are integers $m$ and $n$ so that $9 +\sqrt{11}$ is a root of the polynomial $x^2 + mx + n.$ Find $m + n.$

Let a polynomial $f(x)$ be given with real coefficients and has degree greater or equal than 1. Show that for every real number $c > 0$, there exists a positive integer $n_0$ satisfying the following condition: if polynomial $P(x)$ of degree greater or equal than $n_0$ with real coefficients and has leading coefficient equal to 1 then the number of integers $x$ for which $|f(P(x))| \leq c$ is not greater than degree of $P(x)$.

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.

Provided that the roots of the polynom $ X^n+a_1X^{n-1} +a_2X^{n-2} +\cdots +a_{n-1}X +a_n:\in\mathbb{R}[X] , $ of degree $ n\ge 2, $ are all real and pairwise distinct, prove that there exists is a neighbourhood $ \mathcal{V} $ of $ \left( a_1,a_2,\ldots ,a_n \right) $ in $ \mathbb{R}^n $ and $ n $ functions $ x_1,x_2,\ldots ,x_n\in\mathcal{C}^{\infty } \left( \mathcal{V} \right) $ whose values at $ \left( a_1,a_2,\ldots ,a_n \right) $ are roots of the mentioned polynom.

Determine the coefficients of the equation $$ x^3 - ax^2 + bx - c = 0$$ in such a way that the roots of this equation are the numbers $ a $, $ b $, $ c $.

$p(x)$ is an irreducible polynomial with integer coefficients, and $q$ is a fixed prime number. Let $a_n$ be a number of solutions of the equation $p(x)\equiv 0\mod q^n$. Prove that we can find $M$ such that $\{a_n\}_{n\ge M}$ is constant.

A polynomial $p(x)$ with real coefficients is said to be [i]almeriense[/i] if it is of the form: $$ p(x) = x^3+ax^2+bx+a $$ And its three roots are positive real numbers in arithmetic progression. Find all [i]almeriense[/i] polynomials such that $p\left(\frac{7}{4}\right) = 0$

Find the largest integer $k$ such that $$k\leq\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\sqrt[5]{\frac{5}{4}}+\cdots+\sqrt[2015]{\frac{2015}{2014}}.$$

$P,Q,R$ are non-zero polynomials that for each $z\in\mathbb C$, $P(z)Q(\bar z)=R(z)$. a) If $P,Q,R\in\mathbb R[x]$, prove that $Q$ is constant polynomial. b) Is the above statement correct for $P,Q,R\in\mathbb C[x]$?

A magician intends to perform the following trick. She announces a positive integer $n$, along with $2n$ real numbers $x_1 < \dots < x_{2n}$, to the audience. A member of the audience then secretly chooses a polynomial $P(x)$ of degree $n$ with real coefficients, computes the $2n$ values $P(x_1), \dots , P(x_{2n})$, and writes down these $2n$ values on the blackboard in non-decreasing order. After that the magician announces the secret polynomial to the audience. Can the magician find a strategy to perform such a trick?

Find all $(m,n)$ in $\mathbb{N}^2$ such that $m\mid n^2+1$ and $n\mid m^2+1$.

We define the polynomial $f(x)$ in $\mathbb R[x]$ as follows: $f(x)=x^n+a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+.....+a_1x+a_0$ Prove that there exists an $i$ in the set $\{1,....,n\}$ such that we have $|f(i)|\ge \frac{n!}{\dbinom{n}{i}}$. [i]proposed by Mohammadmahdi Yazdi[/i]

The sequence of polynomials $ \left\{P_n(x)\right\}_{n\equal{}0}^{\plus{}\infty}$ is defined inductively by $ P_0(x) \equal{} 0$ and $ P_{n\plus{}1}(x) \equal{} P_n(x)\plus{}\frac{x \minus{} P_n^2(x)}{2}$. Prove that for any $ x \in [0, 1]$ and any natural number $ n$ it holds that $ 0\le\sqrt x\minus{} P_n(x)\le\frac{2}{n \plus{} 1}$.