[b]p1.[/b] Solve the equation $\log_x (x + 2) = 2$.


[b]p2.[/b] Solve the inequality: $0.5^{|x|} > 0.5^{x^2}$.


[b]p3.[/b] The integers from 1 to 2015 are written on the blackboard. Two randomly chosen numbers are erased and replaced by their difference giving a sequence with one less number. This process is repeated until there is only one number remaining. Is the remaining number even or odd? Justify your answer.


[b]p4.[/b] Four circles are constructed with the sides of a convex quadrilateral as the diameters. Does there exist a point inside the quadrilateral that is not inside the circles? Justify your answer.


[b]p5.[/b] Prove that for any finite sequence of digits there exists an integer the square of which begins with that sequence.


[b]p6.[/b] The distance from the point $P$ to two vertices $A$ and $B$ of an equilateral triangle are $|P A| = 2$ and $|P B| = 3$. Find the greatest possible value of $|P C|$.


PS. You should use hide for answers.
1. We start with the given inequality \(0.5^{|x|} > 0.5^{x^2}\).

2. Since the base \(0.5\) is between 0 and 1, the function \(0.5^y\) is decreasing. Therefore, the inequality \(0.5^{|x|} > 0.5^{x^2}\) implies that \(|x| < x^2\).

3. We analyze the inequality \(|x| < x^2\):
   - For \(x \geq 0\), \(|x| = x\), so the inequality becomes \(x < x^2\).
   - For \(x > 1\), \(x < x^2\) is true because \(x^2 - x > 0\).
   - For \(0 \leq x \leq 1\), \(x < x^2\) is false because \(x^2 - x \leq 0\).
   - For \(x < 0\), \(|x| = -x\), so the inequality becomes \(-x < x^2\).
   - For \(x < -1\), \(-x < x^2\) is true because \(x^2 + x > 0\).
   - For \(-1 \leq x < 0\), \(-x < x^2\) is false because \(x^2 + x \leq 0\).

4. Combining the results, the inequality \(|x| < x^2\) holds for \(x \in (-\infty, -1) \cup (1, \infty)\).

The final answer is \(\boxed{x \in (-\infty, -1) \cup (1, \infty)}\).