[b]p1[/b]. Prove that there is no interger $n$ such that $n^2 +1$ is divisible by $7$.


[b]p2.[/b] Find all solutions of the system
$$x^2-yz=1$$
$$y^2-xz=2$$
$$z^2-xy=3$$


[b]p3.[/b] A triangle with long legs is an isoceles triangle in which the length of the two equal sides is greater than or equal to the length of the remaining side. What is the maximum number, $n$ , of points in the plane with the property that every three of them form the vertices of a triangle with long legs?
Prove all assertions.


[b]p4.[/b] Prove that the area of a quadrilateral of sides $a, b, c, d$ which can be inscribed in a circle and circumscribed about another circle is given by $A=\sqrt{abcd}$


[b]p5.[/b] Prove that all of the squares of side length $$\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},...,\frac{1}{n},...$$ can fit inside a square of side length $1$ without overlap.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. We start with the given system of equations:
   \[
   \begin{cases}
   x^2 - yz = 1 \\
   y^2 - xz = 2 \\
   z^2 - xy = 3
   \end{cases}
   \]

2. Notice the symmetry in the equations. We add all three equations together:
   \[
   x^2 + y^2 + z^2 - xy - yz - xz = 6
   \]

3. Recall the identity:
   \[
   (x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2 - xy - yz - xz)
   \]

4. Using the identity, we multiply the previous equation by 2:
   \[
   2(x^2 + y^2 + z^2 - xy - yz - xz) = 12
   \]

5. This simplifies to:
   \[
   (x - y)^2 + (y - z)^2 + (z - x)^2 = 12
   \]

6. Next, we subtract the original equations from one another. First, subtract equation A from equation B:
   \[
   y^2 - xz - (x^2 - yz) = 2 - 1
   \]
   \[
   y^2 - x^2 + yz - xz = 1
   \]
   \[
   (y - x)(y + x) + (y - x)z = 1
   \]
   \[
   (y - x)(x + y + z) = 1
   \]

7. Similarly, subtract equation B from equation C:
   \[
   z^2 - xy - (y^2 - xz) = 3 - 2
   \]
   \[
   z^2 - y^2 + xz - xy = 1
   \]
   \[
   (z - y)(z + y) + (z - y)x = 1
   \]
   \[
   (z - y)(x + y + z) = 1
   \]

8. Since both equations have the same right-hand side, we subtract them to find relationships between \(x\), \(y\), and \(z\):
   \[
   (z - y)(x + y + z) - (y - x)(x + y + z) = 0
   \]
   \[
   (z - y - y + x)(x + y + z) = 0
   \]
   \[
   (z + x - 2y)(x + y + z) = 0
   \]

9. Since \(x + y + z \neq 0\) (as something multiplied by it is 1), we have:
   \[
   z + x = 2y
   \]

10. Substitute \(z = 2y - x\) into the first equation:
    \[
    (x - y)^2 + (y - (2y - x))^2 + ((2y - x) - x)^2 = 12
    \]
    \[
    (x - y)^2 + (x - y)^2 + (2y - 2x)^2 = 12
    \]
    \[
    2(x - y)^2 + 4(y - x)^2 = 12
    \]
    \[
    6(x - y)^2 = 12
    \]
    \[
    (x - y)^2 = 2
    \]
    \[
    x - y = \pm \sqrt{2}
    \]

11. Let \(x - y = \sqrt{2}\) (we can deal with the negative case later). Then:
    \[
    x = y + \sqrt{2}
    \]

12. Substitute \(x = y + \sqrt{2}\) and \(z = 2y - (y + \sqrt{2}) = y - \sqrt{2}\) into the original equations to find \(y\):
    \[
    (y + \sqrt{2})^2 - y(y - \sqrt{2}) = 1
    \]
    \[
    y^2 + 2y\sqrt{2} + 2 - y^2 + y\sqrt{2} = 1
    \]
    \[
    3y\sqrt{2} + 2 = 1
    \]
    \[
    3y\sqrt{2} = -1
    \]
    \[
    y = -\frac{1}{3\sqrt{2}}
    \]

13. Therefore:
    \[
    x = -\frac{1}{3\sqrt{2}} + \sqrt{2} = \frac{5\sqrt{2}}{6}
    \]
    \[
    z = -\frac{1}{3\sqrt{2}} - \sqrt{2} = -\frac{7\sqrt{2}}{6}
    \]

14. If we used the negative value of \(x - y\) earlier, we also obtain the solutions:
    \[
    x = \frac{5\sqrt{2}}{6}, \quad y = \frac{-\sqrt{2}}{6}, \quad z = \frac{-7\sqrt{2}}{6}
    \]

The final answer is \( \boxed{ x = \frac{5\sqrt{2}}{6}, y = \frac{-\sqrt{2}}{6}, z = \frac{-7\sqrt{2}}{6} } \)