p1. Let $X = \{1, 2, 3, 4, 5\}$. Let $F = \{A_1,A_2,A_3, ..., A_m\}$, with $A_i\subseteq X$ and $2$ members in $A_i$ , for $i = 1,2,...,m$. Determine the minimum $m$ so that for any $B\subseteq X$, with $ B$ having $3$ members, there is a member of $F$ contained in $ B$. Prove your answer.

p2. Find all triples of real numbers $(x, y, z)$ that satisfy the system of equations:

$(x + 1)^2 = x + y + 2$
$(y + 1)^2 = y + z + 2$
$(z + 1)^2 = z + x + 2$

[url=https://artofproblemsolving.com/community/c6h2371594p19389035]p3.[/url] Given the isosceles triangle $ABC$, where $AB = AC$. Let $D$ be a point in the segment $BC$ so that $BD = 2DC$. Suppose also that point $P$ lies on the segment $AD$ such that: $\angle BAC = \angle BP D$. Prove that $\angle BAC = 2\angle DP C$.	

p4. Suppose $p_1, p_2,.., p_n$ arithmetic sequence with difference $b > 0$ and $p_i$ prime for each $i = 1, 2, ..., n$.
a) If $p_1 > n$, prove that every prime number $p$ with $p\le n$, then $p$ divides $b$ .
b)  Give an example of an arithmetic sequence $p_1, p_2,.., p_{10}$, with positive difference and $p_i$ prime for $i = 1, 2, ..., 10$.


p5. Given a set consisting of $22$ integers, $A = \{\pm a_1, \pm a_2, ..., \pm a_{11}\}$. 
Prove that there is a subset  $S$ of $A$ which simultaneously has the following properties:
a) For each $i = 1, 2, ..., 11$ at most only one of the $a_i$ or $-a_i$ is a member of $S$
b) The sum of all the numbers in $S$ is divided by $2015$.
To solve the system of equations:
\[
(x + 1)^2 = x + y + 2
\]
\[
(y + 1)^2 = y + z + 2
\]
\[
(z + 1)^2 = z + x + 2
\]

1. **Sum the equations:**
   \[
   (x + 1)^2 + (y + 1)^2 + (z + 1)^2 = (x + y + 2) + (y + z + 2) + (z + x + 2)
   \]
   Simplifying the right-hand side:
   \[
   (x + y + 2) + (y + z + 2) + (z + x + 2) = 2(x + y + z) + 6
   \]
   Simplifying the left-hand side:
   \[
   (x + 1)^2 + (y + 1)^2 + (z + 1)^2 = x^2 + 2x + 1 + y^2 + 2y + 1 + z^2 + 2z + 1 = x^2 + y^2 + z^2 + 2(x + y + z) + 3
   \]
   Equating both sides:
   \[
   x^2 + y^2 + z^2 + 2(x + y + z) + 3 = 2(x + y + z) + 6
   \]
   Canceling \(2(x + y + z)\) from both sides:
   \[
   x^2 + y^2 + z^2 + 3 = 6
   \]
   Therefore:
   \[
   x^2 + y^2 + z^2 = 3
   \]

2. **Rewrite the first equation:**
   \[
   (x + 1)^2 = x + y + 2 \implies x^2 + 2x + 1 = x + y + 2 \implies x^2 + x - y = 1
   \]
   Similarly, for the second and third equations:
   \[
   y^2 + y - z = 1
   \]
   \[
   z^2 + z - x = 1
   \]

3. **Multiply all three equations:**
   \[
   (x^2 + x - y)(y^2 + y - z)(z^2 + z - x) = 1
   \]
   Notice that if \(x = y = z\), then:
   \[
   x^2 + x - x = 1 \implies x^2 = 1 \implies x = \pm 1
   \]
   Therefore, the possible solutions are \((1, 1, 1)\) and \((-1, -1, -1)\).

4. **Verify the solutions:**
   - For \((1, 1, 1)\):
     \[
     (1 + 1)^2 = 1 + 1 + 2 \implies 4 = 4 \quad \text{(True)}
     \]
     \[
     (1 + 1)^2 = 1 + 1 + 2 \implies 4 = 4 \quad \text{(True)}
     \]
     \[
     (1 + 1)^2 = 1 + 1 + 2 \implies 4 = 4 \quad \text{(True)}
     \]
   - For \((-1, -1, -1)\):
     \[
     (-1 + 1)^2 = -1 - 1 + 2 \implies 0 = 0 \quad \text{(True)}
     \]
     \[
     (-1 + 1)^2 = -1 - 1 + 2 \implies 0 = 0 \quad \text{(True)}
     \]
     \[
     (-1 + 1)^2 = -1 - 1 + 2 \implies 0 = 0 \quad \text{(True)}
     \]

Thus, the solutions are \((1, 1, 1)\) and \((-1, -1, -1)\).

The final answer is \( \boxed{ (1, 1, 1) } \) and \((-1, -1, -1)\).