Found problems: 451
An integer $a$ is called friendly if the equation $(m^2+n)(n^2+m)=a(m-n)^3$ has a solution over the positive integers.
[b]a)[/b] Prove that there are at least $500$ friendly integers in the set $\{ 1,2,\ldots ,2012\}$.
[b]b)[/b] Decide whether $a=2$ is friendly.
How many real solutions are there to the equation $x = 1964 \sin x - 189$ ?
Solve in $ \mathbb{R}^2 $ the equation $ \lfloor x/y-y/x \rfloor =x^2/y+y/x^2. $
Find all rational numbers $a$,for which there exist infinitely many positive rational numbers $q$ such that the equation
$[x^a].{x^a}=q$ has no solution in rational numbers.(A.Vasiliev)
Find the real roots of the equation $x^2 + 2ax + \frac{1}{16} = -a +\sqrt{ a^2 + x - \frac{1}{16} }$ , $\left(0 < a < \frac14 \right)$ .
Show that there are no positive real numbers $x, y, z$ such $(12x^2+yz)(12y^2+xz)(12z^2+xy)= 2014x^2y^2z^2$ .
Prove that $ \forall n > 1, n \in \mathbb{N}$ the equation \[ \sum^n_{k\equal{}1} \frac{x^k}{k!} \plus{} 1 \equal{} 0\] has no rational roots.
Determine, with proof, whether or not there exist integers $a,b,c>2010$ satisfying the equation
\[a^3+2b^3+4c^3=6abc+1.\]
Let $\tau(n)$ denote the number of positive divisors of the positive integer $n$. Prove that there exist infinitely many positive integers $a$ such that the equation $ \tau(an)=n $ does not have a positive integer solution $n$.
Solve in the reals the equation $ 2^{\lfloor\sqrt[3]{x}\rfloor } =x. $
[i]Nedelcu Ion[/i]
Prove that $ \forall n > 1, n \in \mathbb{N}$ the equation \[ \sum^n_{k\equal{}1} \frac{x^k}{k!} \plus{} 1 \equal{} 0\] has no rational roots.
Price of some item has decreased by $5\%$. Then price increased by $40\%$ and now it is $1352.06\$$ cheaper than doubled original price. How much did the item originally cost?
For $a,b, x \in R$ holds: $x^2 - (2a^2 + 4)x + a^2 + 2a + b = 0$. For which $b$ does this equation have at least one root between $0$ and $1$ for all $a$?
The irrational numbers $\alpha ,\beta ,\gamma ,\delta$ are such that $\forall$ $n\in \mathbb{N}$ :
$[n\alpha ].[n\beta ]=[n\gamma ].[n\delta ]$.
Is it true that the sets $\{ \alpha ,\beta \}$ and $\{ \gamma ,\delta \}$ are equal?
Given $I_0 = \{-1,1\}$, define $I_n$ recurrently as the set of solutions $x$ of the equations $x^2 -2xy+y^2- 4^n = 0$,
where $y$ ranges over all elements of $I_{n-1}$. Determine the union of the sets $I_n$ over all nonnegative integers $n$.
Solve the equation $\sqrt{x} = \log_2 x$.
Let $\tau(n)$ denote the number of positive divisors of the positive integer $n$. Prove that there exist infinitely many positive integers $a$ such that the equation $ \tau(an)=n $ does not have a positive integer solution $n$.
Determine all possible values of $x+\frac{1}{x}$ , where the real number $x$ satisfies the equation $$x^4+5x^3-4x^2+5x+1=0$$ and solve this equation.
$(x_n)$ be a sequence with $x_1=0$,
$$x_{n+1}=5x_n + \sqrt{24x_n^2+1}$$.
Prove that for $k \geq 2$ $x_k$ is a natural number.
$(CZS 6)$ Let $d$ and $p$ be two real numbers. Find the first term of an arithmetic progression $a_1, a_2, a_3, \cdots$ with difference $d$ such that $a_1a_2a_3a_4 = p.$ Find the number of solutions in terms of $d$ and $p.$
Solve the equation $\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1$
Show that the equation $x^2 \cdot (x - 1)^2 \cdot (x - 2)^2 \cdot ... \cdot (x - 1008)^2 \cdot (x- 1009)^2 = c$ has $2020$ real solutions, provided $0 < c <\frac{(1009 \cdot1007 \cdot ... \cdot 3\cdot 1)^4}{2^{2020}}$ .
Determine whether there exist distinct real numbers $a, b, c, t$ for which:
[i](i)[/i] the equation $ax^2 + btx + c = 0$ has two distinct real roots $x_1, x_2,$
[i](ii)[/i] the equation $bx^2 + ctx + a = 0$ has two distinct real roots $x_2, x_3,$
[i](iii)[/i] the equation $cx^2 + atx + b = 0$ has two distinct real roots $x_3, x_1.$
Find all positive integers $n$ such that
\[n=d_6^2+d_7^2-1,\]
where $1 = d_1 < d_2 < \cdots < d_k = n$ are all positive divisors of the number $n.$
Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \ldots, a_n$ such that
\[
\frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} =
\frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.
\]
[i]Proposed by Dusan Djukic, Serbia[/i]