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.

Solve the equations : $$\begin{cases} a + b + c = 0 \\ a^2 + b^2 + c^2 = 1\\a^3 + b^3 +c^3 = 4abc \end{cases}$$ for $ a,b,$ and $c. $

Find all triples $(a,b,c)$ of real numbers such that $|2a + b| \ge 4$ and $|ax^2 + bx + c| \le 1$ $ \forall x \in [-1, 1]$.

Prove that $x = y = z = 1$ is the only positive solution of the system \[\left\{ \begin{array}{l} x+y^2 +z^3 = 3\\ y+z^2 +x^3 = 3\\ z+x^2 +y^3 = 3\\ \end{array} \right. \]

Determine all positive real solutions of the following system of equations: $$a =\ max \{ \frac{1}{b} , \frac{1}{c}\} \,\,\,\,\,\, b = \max \{ \frac{1}{c} , \frac{1}{d}\} \,\,\,\,\,\, c = \max \{ \frac{1}{d}, \frac{1}{e}\} $$ $$d = \max \{ \frac{1}{e} , \frac{1}{f }\} \,\,\,\,\,\, e = \max \{ \frac{1}{f} , \frac{1}{a}\} \,\,\,\,\,\, f = \max \{ \frac{1}{a} , \frac{1}{b}\}$$

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.

Find all sets $x,y,z$ of real numbers that satisfy $$\begin{cases} x^3 - y^2 = z^2 - x \\ y^3 -z^2 =x^2 -y \\z^3 -x^2 = y^2 -z \end{cases}$$

Find all triples $(x,y, z)$ of integers such that $$\begin{cases} x^2y + y^2z + z^2x= 2010^2 \\ xy^2 + yz^2 + zx^2= -2010 \end{cases}$$

Prove that if the numbers $ p $, $ q $, $ r $ satisfy the equality $$ p+q + r=1$$ $$ \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0$$ then for any numbers $ a $, $ b $, $ c $ equality holds $$a^2 + b^2 + c^2 = (pa + qb + rc)^2 + (qa + rb + pc)^2 + (ra + pb + qc)^2.$$

Find all triplets $(x, y, z)$ of real numbers for which $$\begin{cases}x^2- yz = |y-z| +1 \\ y^2 - zx = |z-x| +1 \\ z^2 -xy = |x-y| + 1 \end{cases}$$

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

Find all quadruples $(x,y,z,w)$ of integers satisfying the sys­tem of equations $$x + y + z + w = xy + yz + zx + w^2 - w = xyz - w^3 = - 1$$

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.

Determine all real solutions to the system of equations: $$x_1 + x_2 +...+ x_{2004 }= 2004$$ $$x^4_1+ x^4_2+ ... + x^4_{2004} = x^3_1+x^3_2+... + x^3_{2004}$$

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

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 in $R$ the system: $$\begin{cases} \dfrac{xyz}{x + y + 1}= 1998000\\ \\ \dfrac{xyz}{y + z - 1}= 1998000 \\ \\ \dfrac{xyz}{z+x}= 1998000 \end{cases}$$

Solve the following system in positive numbers $\begin{cases} x+y\le 1 \\ \frac{2}{xy} +\frac{1}{x^2+y^2}=10\end{cases}$

On the puzzle page of a newspaper this problem is proposed: “Two children, Antonio and José, have $160$ comics. Antonio counts his by $7$ by $7$ and there are $4$ left over. José counts his $ 8$ by $8$ and he also has $4$ left over. How many comics does he have each?" In the next issue of the newspaper this solution is given: “Antonio has $60$ comics and José has $100$.” Analyze this solution and indicate what a mathematician would do with this problem.

Solve the system of equations $$\begin{cases} (x+y)(x^2-y^2)=32 \\ (x-y)(x^2+y^2)=20 \end{cases}$$

Solve the system of equations $$\begin{cases} \dfrac{1}{x+y}+ x = 3 \\ \\ \dfrac{x}{x+y}=2 \end{cases}$$

Find the smallest positive real $t$ such that \[\left\{ \begin{array}{l} x_1 + x_3 = 2t x_2 \\ x_2 + x_4 = 2t x_3 \\ x_3 + x_5=2t x_4 \\ \end{array} \right. \] has a solution $x_1$, $x_2$, $x_3$, $x_4$, $x_5$ in non-negative reals, not all zero.

Find the values of the real numbers $x,y,z$ that correspond to the system of equations. $$8(x+\frac{1}{x}) =15(y+\frac{1}{y}) = 17(z+\frac{1}{z})$$ $$xy + yz + zx=1$$ [i](Heir of Ramanujan)[/i]

Find all real solutions $x, y$ of the system of equations $$\begin{cases} x + \dfrac{3x-y}{x^2 + y^2} = 3 \\ \\ y-\dfrac{x + 3y}{x^2 + y^2} = 0 \end{cases}$$