[b]p1.[/b] Find all prime numbers $p$ for which $p+10$ and $p+14$ are also prime.


[b]p2.[/b] Prove that $1-\frac12+\frac13-\frac14+...+\frac{1}{1999}-\frac{1}{2000}=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}$.
Can you generalize this formula?


[b]p3.[/b] Suppose somebody has numbered the squares of chessboard by writing in each square one number from $1$ to $64$, without repeating any number. Show that there exist at least two neighboring squares such that the difference between their numbers is at least $5$.  (Note: Two neighboring squares are two squares which have a common side.)


[b]p4.[/b] Suppose we have $101$ points inside a square with one inch sides, placed in such a way that no three points lie on a straight line. Prove  there exist $3$ points such that the triangle they form has area not bigger than $1/100$  in$^2$.



PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
**p1.** Find all prime numbers \( p \) for which \( p+10 \) and \( p+14 \) are also prime.

1. Let \( p \) be a prime number such that \( p+10 \) and \( p+14 \) are also prime.
2. Consider the possible values of \( p \mod 3 \):
   - If \( p \equiv 0 \mod 3 \), then \( p = 3 \). However, \( 3+10 = 13 \) and \( 3+14 = 17 \) are both prime. So, \( p = 3 \) is a solution.
   - If \( p \equiv 1 \mod 3 \), then \( p+10 \equiv 1+10 \equiv 11 \equiv 2 \mod 3 \) and \( p+14 \equiv 1+14 \equiv 15 \equiv 0 \mod 3 \). Since \( p+14 \) is divisible by 3 and greater than 3, it cannot be prime.
   - If \( p \equiv 2 \mod 3 \), then \( p+10 \equiv 2+10 \equiv 12 \equiv 0 \mod 3 \). Since \( p+10 \) is divisible by 3 and greater than 3, it cannot be prime.
3. Therefore, the only solution is \( p = 3 \).

The final answer is \( \boxed{\text{There exist 3 points such that the triangle they form has an area not bigger than } \frac{1}{100} \text{ square inches.}} \).