This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 61

Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1/q$ and a polynomial $P$ with integral coefficients such that \[ \left|P(x)-\frac pq \right| < \frac{1}{q^2}\]for all $x \in I.$
i) A transformation of the plane into itself preserves all rational distances. Prove that it preserves all distances. ii) Show that the corresponding statement for the line is false.
Prove that there is no positive rational number $x$ such that \[x^{\lfloor x\rfloor }=\frac{9}{2}.\]
Prove that for any distinct rational numbers $a, b, c$, the number \[\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}\] is the square of some rational number.
Prove that numbers of the form \[\frac{a_{1}}{1!}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where $0 \le a_{i}\le i-1 \;(i=2, 3, 4, \cdots)$ are rational if and only if starting from some $i$ on all the $a_{i}$'s are either equal to $0$ ( in which case the sum is finite) or all are equal to $i-1$.
Prove that any rational number can be represented as a product of four rational numbers whose sum is zero.
Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1/q$ and a polynomial $P$ with integral coefficients such that \[ \left|P(x)-\frac pq \right| < \frac{1}{q^2}\]for all $x \in I.$
a) Does there exist a real number $x$ such that $x+\sqrt{3}$ and $x^2+\sqrt{3}$ are both rationals? b) Does there exist a real number $y$ such that $y+\sqrt{3}$ and $y^3+\sqrt{3}$ are both rationals?
Let $S$ be a set of rational numbers with the following properties: (a) $\frac12$ is an element in $S$, (b) if $x$ is in $S$, then both $\frac{1}{x+1}$ and $\frac{x}{x+1}$ are in $S$. Prove that $S$ contains all rational numbers in the interval $(0, 1)$.
Let $z=x+yi$ be a complex number with $x$ and $y$ rational and with $|z|=1.$ Prove that the number $|z^{2n} -1|$ is rational for every integer $n$.