Find all fourth degree polynomials $p(x)$ such that the following four conditions are satisfied:

(i) $p(x)=p(-x)$ for all $x$.

(ii) $p(x) \geq 0$ for all $x$.

(iii) $p(0)=1$.

(iv) $p(x)$ has exactly two local minimum points $x_{1}$ and $x_{2}$ such that $\left|x_{1}-x_{2}\right|=2$.
Let $p(x)=a x^{4}+b x^{3}+c x^{2}+d x+e$ with $a \neq 0$. From $(i)-(i i i)$ we get $b=d=0, a>0$ and $e=1$. From (iv) it follows that $p^{\prime}(x)=4 a x^{3}+2 c x$ has at least two different real roots. Since $a>0$, we have $c<0$ and $p^{\prime}(x)$ has three roots $x=0, x= \pm \sqrt{-c /(2 a)}$. The minimum points mentioned in $(i v)$ must be $x= \pm \sqrt{-c /(2 a)}$, so $2 \sqrt{-c /(2 a)}=2$ and $c=-2 a$. Finally, by $(i i)$ we have $p(x)=a\left(x^{2}-1\right)^{2}+1-a \geq 0$ for all $x$, which implies $0<a \leq 1$. It is easy to check that every such polynomial satisfies the conditions (i)-(iv).