Each integer is colored with exactly one of $3$ possible colors -- black, red or white -- satisfying the following two rules : the negative of a black number must be colored white, and the sum of two white numbers (not necessarily distinct) must be colored black.

[b](a)[/b] Show that, the negative of a white number must be colored black and the sum of two black numbers must be colored white.

[b](b)[/b] Determine all possible colorings of the integers that satisfy these rules.
1. **Show that the negative of a white number must be colored black:**

   Suppose \( a \) is white. According to the given rules:
   - The sum of two white numbers must be black. Therefore, \( a + a = 2a \) is black.
   - The negative of a black number must be white. Therefore, \( -2a \) is white.
   - Now consider \( a - 2a = -a \). Since \( a \) is white and \( -2a \) is white, their sum \( -a \) must be black.

   Hence, if \( a \) is white, then \( -a \) must be black.

2. **Show that the sum of two black numbers must be colored white:**

   Suppose \( a \) and \( b \) are black. According to the given rules:
   - The negative of a black number must be white. Therefore, \( -a \) and \( -b \) are white.
   - The sum of two white numbers must be black. Therefore, \( -a + (-b) = -(a + b) \) is black.
   - The negative of a black number must be white. Therefore, \( a + b \) must be white.

   Hence, the sum of two black numbers must be white.

3. **Determine all possible colorings of the integers that satisfy these rules:**

   Let's analyze the possible colorings:
   - From the rules, we know that if \( a \) is white, then \( -a \) is black.
   - If \( a \) is black, then \( -a \) is white.
   - The sum of two white numbers is black.
   - The sum of two black numbers is white.

   Consider the number \( 0 \):
   - If \( 0 \) is white, then \( 0 + 0 = 0 \) must be black, which is a contradiction.
   - If \( 0 \) is black, then \( 0 + 0 = 0 \) must be white, which is also a contradiction.

   Therefore, \( 0 \) cannot be colored either black or white. This implies that \( 0 \) must be the third color, red.

   Now, consider the implications:
   - If \( a \) is any non-zero integer, it must be either black or white.
   - If \( a \) is white, then \( -a \) is black.
   - If \( a \) is black, then \( -a \) is white.

   This leads to the conclusion that all non-zero integers must be either black or white, and \( 0 \) must be red.

   Therefore, the only possible coloring that satisfies all the given rules is:
   - \( 0 \) is red.
   - All positive integers are either black or white.
   - All negative integers are colored such that the negative of a black number is white and vice versa.

The final answer is \( \boxed{ 0 } \) is red, and all non-zero integers are either black or white, with the negative of a black number being white and vice versa.