Olympiad problems

2010 N.N. Mihăileanu Individual, 1

Let be two real reducible quadratic polynomials $ P,Q $ in one variable. Prove that if $ P-Q $ is irreducible, then $ P+Q $ is reducible.

2013 AMC 10, 19

Tags: quadratics , AMC , AMC 10 , AMC 10 B , Vieta , arithmetic sequence , algebra

The real numbers $c, b, a$ form an arithmetic sequence with $a\ge b\ge c\ge 0$. The quadratic $ax^2+bx+c$ has exactly one root. What is this root? $\textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} $

1962 Miklós Schweitzer, 4

Tags: floor function , modular arithmetic , quadratics , symmetry , advanced fields , advanced fields unsolved

Show that \[ \prod_{1\leq x < y \leq \frac{p\minus{}1}{2}} (x^2\plus{}y^2) \equiv (\minus{}1)^{\lfloor\frac{p\plus{}1}{8}\rfloor} \;(\textbf{mod}\;p\ ) \] for every prime $ p\equiv 3 \;(\textbf{mod}\;4\ )$. [J. Suranyi]

2003 Vietnam Team Selection Test, 3

Tags: modular arithmetic , quadratics , number theory unsolved , number theory

Let $n$ be a positive integer. Prove that the number $2^n + 1$ has no prime divisor of the form $8 \cdot k - 1$, where $k$ is a positive integer.

2005 AIME Problems, 7

Tags: trigonometry , quadratics , algebra , quadratic formula , trig identities , Law of Cosines

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

2013 International Zhautykov Olympiad, 1

Tags: quadratics , number theory , relatively prime , algebra unsolved , algebra

A quadratic trinomial $p(x)$ with real coefficients is given. Prove that there is a positive integer $n$ such that the equation $p(x) = \frac{1}{n}$ has no rational roots.

1989 IMO Shortlist, 25

Tags: number theory , Diophantine equation , quadratics , equation , IMO Shortlist

Let $ a, b \in \mathbb{Z}$ which are not perfect squares. Prove that if \[ x^2 \minus{} ay^2 \minus{} bz^2 \plus{} abw^2 \equal{} 0\] has a nontrivial solution in integers, then so does \[ x^2 \minus{} ay^2 \minus{} bz^2 \equal{} 0.\]