Prove that, if $a$ and $b$ are given integers, the system of equatìons $$x + y + 2z + 2t = a$$ $$2x - 2y + z- t = b$$ has a solution in integers $x, y,z,t$.

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 $n > 1$ 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.

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 minimal value of $a \in \mathbb{R}$ such that system $$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a-1$$ $$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$$ has solution in set of real numbers

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.

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.

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

Solve the system of equations $$ \left\{\begin{array}{l} x \log x+y \log y+z \log x=0\\ \\ \dfrac{\log x}{x}+\dfrac{\log y}{y}+\dfrac{\log z}{z}=0 \end{array} \right. $$

Let $x, y,z$ satisfy the following inequalities $\begin{cases} | x + 2y - 3z| \le 6 \\ | x - 2y + 3z| \le 6 \\ | x - 2y - 3z| \le 6 \\ | x + 2y + 3z| \le 6 \end{cases}$ Determine the greatest value of $M = |x| + |y| + |z|$.

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. $

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.

Find all triples $(x,y,z)$ of positive integers such that $$\begin{cases} x + y +z = 2010 \\x^2 + y^2 + z^2 - xy - yz - zx =3 \end{cases}$$

Solve in $N$: $$\begin{cases} a^3=b^3+c^3+12a \\ a^2=5(b+c) \end{cases}$$

A system of equations $$\begin{cases} x^2 \,\, ? \,\, z^2 = -8 \\ y^2 \,\, ? \,\, z^2 = 7 \end{cases}$$ is written on a piece of paper, but unfortunately two of the symbols are a little blurred. However, it is known that the system has at least one solution, and that each of the two question marks stands for either $+$ or $-$. What are the two symbols?

Show that the only integral solution to \[\left\{ \begin{array}{l} xy + yz + zx = 3n^2 - 1\\ x + y + z = 3n \\ \end{array} \right. \] with $x \geq y \geq z$ is $x=n+1$, $y=n$, $z=n-1$.

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 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, for any positive real number $a$, the number of solutions $(x,y)$ to the system of equations $$\begin{cases} |x|+|y|= 1 \\ x^2 + y^2 = a \end{cases}$$ where $x$ and $y$ are real numbers.

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

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: $$x_1 + x_2 +...+ x_{2004 }= 2004$$ $$x^4_1+ x^4_2+ ... + x^4_{2004} = x^3_1+x^3_2+... + x^3_{2004}$$

Real numbers $a$ and $b$ satisfy the system of equations $$\begin{cases} a^3-a^2+a-5=0 \\ b^3-2b^2+2b+4=0 \end{cases}$$ Find the numerical value of the sum $a+ b$.

The real numbers x and у satisfy the equations $$\begin{cases} \sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2 \\ \\ \sqrt{7y}\left(1-\dfrac{1}{x+y}\right)=4\sqrt2 \end{cases}$$ Find the numerical value of the ratio $y/x$.

Solve a set of inequalities in the domain of integer numbers: $$3x^2 +2yz \le 1+y^2$$ $$3y^2 +2zx \le 1+z^2$$ $$3z^2 +2xy \le 1+x^2$$