已知 $y = x^3 + px^2 + qx + 10$，其中 $p$ 和 $q$ 是整数。使 $y$ 为 $x$ 的递减函数的 $x$ 值的唯一区间为 $3 < x < 7$。求 $p$ 和 $q$ 的值。

1. Find the derivative of $y$ with respect to $x$.

$\frac{dy}{dx} = 3x^2 + 2px + q$

2. Relate the decreasing condition to the roots of the derivative.

For $y$ to be a decreasing function, $\frac{dy}{dx} < 0$.
Since $\frac{dy}{dx} < 0$ only for $3 < x < 7$, the roots of the quadratic equation $\frac{dy}{dx} = 0$ must be $x = 3$ and $x = 7$.

3. Form the quadratic expression using the roots and compare coefficients.

$\frac{dy}{dx} = 3(x - 3)(x - 7)$
$= 3(x^2 - 10x + 21)$
$= 3x^2 - 30x + 63$

Comparing coefficients with $3x^2 + 2px + q$:
$2p = -30 \implies p = -15$
$q = 63$