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 $$0 <x^2 +y^2 < \delta^2.$$

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 $\textbf{(A) }\frac{1}{256}\qquad\textbf{(B) }-\frac{1}{16}\qquad\textbf{(C) }1\qquad\textbf{(D) }-16\qquad\textbf{(E) }256$

A [i]self-avoiding rook walk[/i] on a chessboard (a rectangular grid of unit squares) is a path traced by a sequence of moves parallel to an edge of the board from one unit square to another, such that each begins where the previous move ended and such that no move ever crosses a square that has previously been crossed, i.e., the rook's path is non-self-intersecting. Let $ R(m, n)$ be the number of self-avoiding rook walks on an $ m \times n$ ($ m$ rows, $ n$ columns) chessboard which begin at the lower-left corner and end at the upper-left corner. For example, $ R(m, 1) \equal{} 1$ for all natural numbers $ m$; $ R(2, 2) \equal{} 2$; $ R(3, 2) \equal{} 4$; $ R(3, 3) \equal{} 11$. Find a formula for $ R(3, n)$ for each natural number $ n$.

Let $n$ be a positive integer, $h$ a real number and $f(x)$ a polynomial (whole rational function) with real coefficients of degree n, which has no real zeros. Prove that then also the polynomial $$F(x) = f(x) + h f'(x) + h^2 f''(x) +... + h^n f^{(n)}(x)$$ has no real zeros.

Let $n$ be a positive integer. Solve the system of equations \begin{align*}x_{1}+2x_{2}+\cdots+nx_{n}&= \frac{n(n+1)}{2}\\ x_{1}+x_{2}^{2}+\cdots+x_{n}^{n}&= n\end{align*} for $n$-tuples $(x_{1},x_{2},\ldots,x_{n})$ of nonnegative real numbers.

How many polynomials $P$ of integer coefficients and degree at most $4$ satisfy $0 \le P(x) < 72$ for all $x\in \{0, 1, 2, 3, 4\}$? Harvard-MIT Mathematics Tournament 2011

Suppose $P(x)$ is a quadratic polynomial with integer coefficients satisfying the identity \[P(P(x)) - P(x)^2 = x^2+x+2016\] for all real $x$. What is $P(1)$?

[b]p1.[/b] A band of pirates unloaded some number of treasure chests from their ship. The number of pirates was between $60$ and $69$ (inclusive). Each pirate handled exactly $11$ treasure chests, and each treasure chest was handled by exactly $7$ pirates. Exactly how many treasure chests were there? Show that your answer is the only solution. [b]p2.[/b] Let $a$ and $b$ be the lengths of the legs of a right triangle, let $c$ be the length of the hypotenuse, and let $h$ be the length of the altitude drawn from the vertex of the right angle to the hypotenuse. Prove that $c+h>a+b$. [b]p3.[/b] Prove that $$\frac{1}{70}< \frac{1}{2} \frac{3}{4} \frac{5}{6} ... \frac{2001}{2002} < \frac{1}{40}$$ [b]p4.[/b] Given a positive integer $a_1$ we form a sequence $a_1 , a_2 , a _3,...$ as follows: $a_2$ is obtained from $a_1$ by adding together the digits of $a_1$ raised to the $2001$-st power; $a_3$ is obtained from $a_2$ using the same rule, and so on. For example, if $a_1 =25$, then $a_2 =2^{2001}+5^{2001}$, which is a $1399$-digit number containing $106$ $0$'s, $150$ $1$'s, 4124$ 42$'s, $157$ $3$'s, $148$ $4$'s, $141$ $5$'s, $128$ $6$'s, $1504 47$'s, $152$ $8$'s, $143$ $9$'s. So $a_3 = 106 \times 0^{2001}+ 150 \times 1^{2001}+ 124 \times 2^{2001}+ 157 \times 3^{2001}+ ...+ 143 \times 9^{2001}$ which is a $1912$-digit number, and so forth. Prove that if any positive integer $a_1$ is chosen to start the sequence, then there is a positive integer $M$ (which depends on $a_1$ ) that is so large that $a_n < M$ for all $n=1,2,3,...$ [b]p5.[/b] Let $P(x)$ be a polynomial with integer coefficients. Suppose that there are integers $a$, $b$, and $c$ such that $P(a)=0$, $P(b)=1$, and $P(c)=2$. Prove that there is at most one integer $n$ such that $P(n)=4$. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Consider (3-variable) polynomials \[P_n(x,y,z)=(x-y)^{2n}(y-z)^{2n}+(y-z)^{2n}(z-x)^{2n}+(z-x)^{2n}(x-y)^{2n}\] and \[Q_n(x,y,z)=[(x-y)^{2n}+(y-z)^{2n}+(z-x)^{2n}]^{2n}.\] Determine all positive integers $n$ such that the quotient $Q_n(x,y,z)/P_n(x,y,z)$ is a (3-variable) polynomial with rational coefficients.

Let $g(x)$ be a fixed polynomial with real coefficients and define $f(x)$ by $f(x) =x^2 + xg(x^3)$. Show that $f(x)$ is not divisible by $x^2 - x + 1$.

Let $P(x)$ be a polynomial with integer coefficients and roots $1997$ and $2010$. Suppose further that $|P(2005)|<10$. Determine what integer values $P(2005)$ can get.

Let $f(x) = x^m + a_1x^{m-1} + \cdots+ a_{m-1}x + a_m$ and $g(x) = x^n + b_1x^{n-1} + \cdots + b_{n-1}x + b_n$ be two polynomials with real coefficients such that for each real number $x, f(x)$ is the square of an integer if and only if so is $g(x)$. Prove that if $n +m > 0$, then there exists a polynomial $h(x)$ with real coefficients such that $f(x) \cdot g(x) = (h(x))^2.$ [hide="Remark."]Remark. The original problem stated $g(x) = x^n + b_1x^{n-1} + \cdots + {\color{red}{ b_{n-1}}} + b_n$, but I think the right form of the problem is what I wrote.[/hide]

Determine all polynomials $P(x)$ with integer coefficients such that, for any positive integer $n$, the equation $P(x)=2^n$ has an integer root.

Determine all integers $n$ such that both of the numbers: $$|n^3 - 4n^2 + 3n - 35| \text{ and } |n^2 + 4n + 8|$$ are both prime numbers.

Find the sum of all positive integers $1\leq k\leq 99$ such that there exist positive integers $a$ and $b$ with the property that \[x^{100}-ax^k+b=(x^2-2x+1)P(x)\] for some polynomial $P$ with integer coefficients. [i]Proposed by David Altizio[/i]