一元二次方程 $x^{2} + px + 4 = 0$ 的根为 $\alpha^{2}$ 和 $\beta^{2}$，其中 $\alpha > 0$，$\beta > 0$ 且 $\alpha > \beta$。

1. Find the sum and product of the roots $\alpha^2$ and $\beta^2$.

Sum of roots: $\alpha^2 + \beta^2 = -p$
Product of roots: $\alpha^2 \beta^2 = 4$
Since $\alpha > 0$ and $\beta > 0$, $\alpha\beta = \sqrt{4} = 2$.

2. Express $(\alpha - \beta)^2$ in terms of $p$.

$(\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2$
$(\alpha - \beta)^2 = (\alpha^2 + \beta^2) - 2\alpha\beta$
$(\alpha - \beta)^2 = -p - 2(2) = -p - 4$

3. Find $\alpha - \beta$ by taking the square root.

Since $\alpha > \beta$, $\alpha - \beta > 0$.
Therefore, $\alpha - \beta = \sqrt{-p - 4}$.

1. Substitute $p = -5$ into the original equation and solve for $\alpha^2$ and $\beta^2$.

When $p = -5$, the equation is $x^{2} - 5x + 4 = 0$.
$(x - 4)(x - 1) = 0$
$x = 4 \text{ or } x = 1$
Since the roots are $\alpha^2$ and $\beta^2$, the values are $4$ and $1$.

2. Determine the exact values of $\alpha$ and $\beta$ using the given conditions.

Given $\alpha > \beta$, we must have $\alpha^2 = 4$ and $\beta^2 = 1$.
Since $\alpha > 0$ and $\beta > 0$:
$\alpha = \sqrt{4} = 2$
$\beta = \sqrt{1} = 1$

3. Calculate the roots of the new quadratic equation.

The new roots are $\alpha - 1$ and $1 - \beta$.
$\alpha - 1 = 2 - 1 = 1$
$1 - \beta = 1 - 1 = 0$

4. Form the new quadratic equation.

Sum of new roots $= 1 + 0 = 1$
Product of new roots $= 1 \times 0 = 0$
The new quadratic equation is $x^{2} - (\text{sum})x + (\text{product}) = 0$:
$x^{2} - x = 0$