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The parabolas

\begin{aligned} y & = a + b x + x^{2} \\ y & = b + a x - x^{2}, \end{aligned}

where $a, b \in Z,$ do not intersect. For each value of $a$ , how many distinct values could $b$ take?

Hint

The quadratic

a + b x + x^{2} = b + a x - x^{2}

must have negative discriminant.

Hint

View your discriminant as a quadratic in terms of $b$ - imagine $a$ is a fixed constant.

In your sketch, $b$ should be on the horizontal axis.

Solution

Since the curves do not intersect, we require the discriminant of the equation

\begin{aligned} a + b x + x^{2} & = b + a x - x^{2} \\ 2 x^{2} + (b - a) x + a - b & = 0 \end{aligned}

to be negative, so

\begin{aligned} (b - a)^{2} - 4 (2) (a - b) & < 0 \\ (b - a)^{2} - 8 (a - b) & < 0 \\ (b - a)^{2} + 8 (b - a) & < 0 \\ (b - a) (b - a + 8) & < 0 \end{aligned}

Viewing this as a quadratic in terms of $b$ , we note that the roots are

\begin{aligned} b & = a \\ b & = a - 8 \end{aligned}

This enables us to make a sketch

parabola

Since $a \in Z$ , we can be confident that there are $7$ values of $b \in Z$ between $a - 8$ and $a$ , thus making the discriminant negative and guaranteeing no intersections.