Find the real numbers $ x, y, z $ satisfying the system of equations $$(z - x)(x - y) = a $$ $$(x - y)(y - z) = b$$ $$(y - z)(z - x) = c$$ where $ a, b, c $ are given real numbers.

Let $n > 1$ be an integer. In a circular arrangement of $n$ lamps $L_0, \ldots, L_{n-1},$ each of of which can either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $Step_0, Step_1, \ldots .$ If $L_{j-1}$ ($j$ is taken mod $n$) is ON then $Step_j$ changes the state of $L_j$ (it goes from ON to OFF or from OFF to ON) but does not change the state of any of the other lamps. If $L_{j-1}$ is OFF then $Step_j$ does not change anything at all. Show that: (i) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again, (ii) If $n$ has the form $2^k$ then all the lamps are ON after $n^2-1$ steps, (iii) If $n$ has the form $2^k + 1$ then all lamps are ON after $n^2 - n + 1$ steps.

Find all pairs $(x, y)$ of real numbers that satisfy the system of equations $$\begin{cases} x^4 + 2z^3 - y =\sqrt3 - \dfrac14 \\ y^4 + 2y^3 - x = - \sqrt3 - \dfrac14 \end{cases}$$

Let $a > 0$. If the system $$\begin{cases} a^x + a^y + a^z = 14 - a \\ x + y + z = 1 \end{cases}$$ has a solution in real numbers, prove that $a \le 8$.

Solve \[\left\{ \begin{array}{l} y(x+y)^2 = 9 \\ y(x^3-y^3) = 7 \\ \end{array} \right. \]

Let $p$ be a prime number. The natural numbers $m$ and $n$ are written in the system with the base $p$ as $n = a_0 + a_1p +...+ a_kp^k$ and $m = b_0 + b_1p +..+ b_kp^k$. Prove that $${n \choose m} \equiv \prod_{i=0}^{k}{a_i \choose b_i} (mod p)$$

Given $k_1,k_2,...,k_n\in R^+$, find all the naturals $n$ such that $$k_1+k_2+...+k_n=2n-3$$ $$\frac{1}{k_1}+\frac{1}{k_2}+...+\frac{1}{k_n}=3$$ [i](Zhuge Liang)[/i]

Find all triples $(a, b, c)$ of positive integers for which $$\begin{cases} a + bc=2010 \\ b + ca = 250\end{cases}$$

Determine all real solutions to the system of equations: $$x^2 - y = z^2$$ $$y^2 - z = x^2$$ $$z^2 - x = y^2$$

Determine all sets of real numbers $(x,y,z)$ which fulfills $$\begin{cases} x + y =2 \\ xy -z^2= 1\end{cases}$$

Solve in natural numbers $a,b,c$ the system \[\left\{ \begin{array}{l}a^3 -b^3 -c^3 = 3abc \\ a^2 = 2(a+b+c)\\ \end{array} \right. \]

Let $n$ be an integer $> 1$. In a circular arrangement of $n$ lamps $L_0, \cdots, L_{n-1}$, each one of which can be either ON or OFF, we start with the situation that all lamps are ON, and then carry out a sequence of steps, $Step_0, Step_1, \cdots$. If $L_{j-1}$ ($j$ is taken mod n) is ON, then $Step_j$ changes the status of $L_j$ (it goes from ON to OFF or from OFF to ON) but does not change the status of any of the other lamps. If $L_{j-1}$ is OFF, then $Step_j$ does not change anything at all. Show that: [i](a)[/i] There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again. [i](b)[/i] If $n$ has the form $2^k$, then all lamps are ON after $n^2 - 1$ steps. [i](c) [/i]If $n$ has the form $2^k +1$, then all lamps are ON after $n^2 -n+1$ steps.

Solve the system of equations: $$|\log_2(x + y)| + | \log_2(x - y)| = 3$$ $$xy = 3$$

Find all quadruplets $(x_1, x_2, x_3, x_4)$ of real numbers such that the next six equalities apply: $$\begin{cases} x_1 + x_2 = x^2_3 + x^2_4 + 6x_3x_4\\ x_1 + x_3 = x^2_2 + x^2_4 + 6x_2x_4\\ x_1 + x_4 = x^2_2 + x^2_3 + 6x_2x_3\\ x_2 + x_3 = x^2_1 + x^2_4 + 6x_1x_4\\ x_2 + x_4 = x^2_1 + x^2_3 + 6x_1x_3 \\ x_3 + x_4 = x^2_1 + x^2_2 + 6x_1x_2 \end{cases}$$

If $x, y, z, w$ are nonnegative real numbers satisfying \[\left\{ \begin{array}{l}y = x - 2003 \\ z = 2y - 2003 \\ w = 3z - 2003 \\ \end{array} \right. \] find the smallest possible value of $x$ and the values of $y, z, w$ corresponding to it.