p1. Calculate all the solutions $(m, n)$ of integers that satisfy the equation $mn^2 = 2009 (n + 1)$.


p2. A cube is cut from side to by a plane that passes through the midpoints of three edges that are on different faces. Determine the plane figure that appears in the cut (the section) and calculate its area.


p3. Prove that   $$\frac{x}{1 + x^2} +\frac{y}{1 + x^2 + y^2} <1$$ for any choice of real numbers $x, y$.


p4. Prove that every $n$-sided polygon (not necessarily convex) has at least three interior angles that are less than $180$ degrees.


p5. Is it possible to divide the numbers from $1$ to $100$ into fifty ordered pairs
$$(a_1, b_1), (a_2, b_2),..., (a_{50}, b_{50})$$
so in the list
$$|a_1-b_1|, |a_2-b_2|, ... , |a_{50}- b_{50}|$$
$$|a_1- a_2|, |a_2-a_3|,  ... , |a_{49}- a_{50}|$$
all integers from $ 1$ to 9$9$ are found?


p6. Consider $m, n$ positive integers with $m> n$. Determine if the integer part of the number $\frac{2^m-1}{2^n -1} $ is odd or even.

PS. Seniors P1, P2 were also proposed as [url=https://artofproblemsolving.com/community/c4h2690183p23348577]Juniors[/url] P1, P2.
1. Start with the given equation:
   \[
   mn^2 = 2009(n + 1)
   \]
   Rearrange the equation to form a quadratic equation in \( n \):
   \[
   mn^2 - 2009n - 2009 = 0
   \]

2. Recognize that this is a quadratic equation in \( n \) of the form \( an^2 + bn + c = 0 \), where \( a = m \), \( b = -2009 \), and \( c = -2009 \). For \( n \) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \( \Delta \) is given by:
   \[
   \Delta = b^2 - 4ac = (-2009)^2 - 4 \cdot m \cdot (-2009)
   \]
   Simplify the discriminant:
   \[
   \Delta = 2009^2 + 4 \cdot 2009 \cdot m = 2009(2009 + 4m)
   \]

3. For \( n \) to be an integer, \( \Delta \) must be a perfect square. Let \( k^2 = 2009(2009 + 4m) \) for some integer \( k \). This implies:
   \[
   2009(2009 + 4m) = k^2
   \]
   Rearrange to solve for \( m \):
   \[
   2009 + 4m = \frac{k^2}{2009}
   \]
   \[
   4m = \frac{k^2}{2009} - 2009
   \]
   \[
   m = \frac{k^2 - 2009^2}{4 \cdot 2009}
   \]

4. For \( m \) to be an integer, \( k^2 - 2009^2 \) must be divisible by \( 4 \cdot 2009 \). Let \( k = 2009 + 2t \) for some integer \( t \). Then:
   \[
   k^2 = (2009 + 2t)^2 = 2009^2 + 4 \cdot 2009 \cdot t + 4t^2
   \]
   \[
   k^2 - 2009^2 = 4 \cdot 2009 \cdot t + 4t^2
   \]
   \[
   m = \frac{4 \cdot 2009 \cdot t + 4t^2}{4 \cdot 2009} = t + \frac{t^2}{2009}
   \]

5. For \( m \) to be an integer, \( \frac{t^2}{2009} \) must be an integer. This implies \( t^2 \) must be a multiple of 2009. Since 2009 is a prime number, \( t \) must be a multiple of 2009. Let \( t = 2009k \) for some integer \( k \). Then:
   \[
   m = 2009k(k + 1)
   \]

6. Substitute \( m = 2009k(k + 1) \) back into the quadratic equation and solve for \( n \):
   \[
   2009k(k + 1)n^2 - 2009n - 2009 = 0
   \]
   Divide through by 2009:
   \[
   k(k + 1)n^2 - n - 1 = 0
   \]
   Solve this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   n = \frac{1 \pm \sqrt{1 + 4k(k + 1)}}{2k(k + 1)}
   \]

7. For \( n \) to be an integer, the discriminant \( 1 + 4k(k + 1) \) must be a perfect square. Let \( 1 + 4k(k + 1) = m^2 \) for some integer \( m \):
   \[
   m^2 - 1 = 4k(k + 1)
   \]
   \[
   (m - 1)(m + 1) = 4k(k + 1)
   \]

8. Solve for \( k \) and \( m \):
   - If \( k = 1 \), then \( m = 3 \):
     \[
     (m, n) = (2009 \cdot 1 \cdot 2, 1) = (4018, 1)
     \]
   - If \( k = 0 \), then \( m = 1 \):
     \[
     (m, n) = (0, -1)
     \]

9. Therefore, the solutions are:
   \[
   (m, n) = (4018, 1) \quad \text{and} \quad (m, n) = (0, -1)
   \]

The final answer is \(\boxed{(4018, 1), (0, -1)}\)