Find all fourth degree polynomial $ p(x)$ such that the following four conditions are satisfied:
(i) $ p(x)\equal{}p(\minus{}x)$ for all $ x$,
(ii) $ p(x)\ge0$ for all $ x$,
(iii) $ p(0)\equal{}1$
(iv) $ p(x)$ has exactly two local minimum points $ x_1$ and $ x_2$ such that $ |x_1\minus{}x_2|\equal{}2$.
1. **Symmetry Condition**: Given that \( p(x) = p(-x) \) for all \( x \), \( p(x) \) must be an even function. Therefore, \( p(x) \) can be expressed as:
   \[
   p(x) = ax^4 + bx^2 + c
   \]
   where \( a \), \( b \), and \( c \) are constants.

2. **Non-negativity Condition**: Given that \( p(x) \ge 0 \) for all \( x \), the polynomial must be non-negative for all real numbers \( x \).

3. **Value at Zero**: Given that \( p(0) = 1 \), substituting \( x = 0 \) into the polynomial gives:
   \[
   p(0) = c = 1
   \]
   Therefore, the polynomial can be rewritten as:
   \[
   p(x) = ax^4 + bx^2 + 1
   \]

4. **Local Minima Condition**: Given that \( p(x) \) has exactly two local minimum points \( x_1 \) and \( x_2 \) such that \( |x_1 - x_2| = 2 \), we need to find the critical points of \( p(x) \). The critical points are found by setting the first derivative \( p'(x) \) to zero:
   \[
   p'(x) = 4ax^3 + 2bx = 0
   \]
   Factoring out \( x \) gives:
   \[
   x(4ax^2 + 2b) = 0
   \]
   This equation has solutions:
   \[
   x = 0 \quad \text{or} \quad 4ax^2 + 2b = 0
   \]
   Solving \( 4ax^2 + 2b = 0 \) for \( x \) gives:
   \[
   4ax^2 = -2b \quad \Rightarrow \quad x^2 = -\frac{b}{2a}
   \]
   For \( x \) to be real, \( -\frac{b}{2a} \) must be non-negative, implying \( b \le 0 \).

5. **Distance Between Minima**: Given \( |x_1 - x_2| = 2 \), and knowing that \( x_1 \) and \( x_2 \) are symmetric about the origin (since \( p(x) \) is even), we can assume \( x_1 = 1 \) and \( x_2 = -1 \). Therefore:
   \[
   1^2 = -\frac{b}{2a} \quad \Rightarrow \quad 1 = -\frac{b}{2a} \quad \Rightarrow \quad b = -2a
   \]

6. **Substitute \( b \) into \( p(x) \)**: Substituting \( b = -2a \) into the polynomial gives:
   \[
   p(x) = ax^4 - 2ax^2 + 1
   \]

7. **Non-negativity Check**: To ensure \( p(x) \ge 0 \) for all \( x \), we rewrite \( p(x) \) as:
   \[
   p(x) = a(x^4 - 2x^2) + 1 = a(x^2 - 1)^2 + 1 - a
   \]
   For \( p(x) \ge 0 \), the minimum value of \( a(x^2 - 1)^2 \) is 0 (when \( x = \pm 1 \)), so:
   \[
   1 - a \ge 0 \quad \Rightarrow \quad a \le 1
   \]
   Since \( a > 0 \) (to ensure the polynomial is of degree 4 and non-negative), we have:
   \[
   0 < a \le 1
   \]

Conclusion:
The fourth-degree polynomial \( p(x) \) that satisfies all the given conditions is:
\[
p(x) = a(x^2 - 1)^2 + 1 - a
\]
where \( 0 < a \le 1 \).

The final answer is \( \boxed{ p(x) = a(x^2 - 1)^2 + 1 - a } \) where \( 0 < a \le 1 \).