已知曲线 $y = ax^2 + 2bx - 1$ 完全位于 x 轴下方，求常数 $a$ 和 $b$ 必须满足的条件。 如果 $a$ 和 $b$ 均为整数，写出一组满足在 (i) 中求得的条件的 $a$ 和 $b$ 的值。

1. State the condition for the curve to open downwards

For the curve to lie entirely below the x-axis, it must open downwards, so the coefficient of $x^2$ must be negative:
$a < 0$

2. Apply the discriminant condition for no real roots

The curve must not intersect the x-axis, so the discriminant must be less than zero:
$b^2 - 4ac < 0$
$(2b)^2 - 4(a)(-1) < 0$
$4b^2 + 4a < 0$
$b^2 + a < 0$
$a < -b^2$

1. Recall the conditions from part (i)

From (i), the conditions are $a < 0$ and $b^2 + a < 0$ (or $a < -b^2$).

2. Choose integer values for $a$ and $b$ that satisfy the conditions

Let $b = 1$.
Then $a < -(1)^2 \implies a < -1$.
Choosing an integer $a$ that satisfies this, we can let $a = -2$.