HMMT November

Algebra Difficulty 4.9 AIME Find the answer

Determine the number of integers $D$ such that whenever $a$ and $b$ are both real numbers with $-1 / 4<a, b<1 / 4$ , then $\left|a^{2}-D b^{2}\right|<1$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have $-1<a^{2}-D b^{2}<1 \Rightarrow \frac{a^{2}-1}{b^{2}}<D<\frac{a^{2}+1}{b^{2}}$ We have $\frac{a^{2}-1}{b^{2}}$ is maximal at $-15=\frac{.25^{2}-1}{.25^{2}}$ and $\frac{a^{2}+1}{b^{2}}$ is minimal at $\frac{0^{2}+1}{.25^{2}}=16$ . However, since we cannot have $a, b= \pm .25$ , checking border cases of -15 and 16 shows that both of these values are possible for $D$ . Hence, $-15 \leq D \leq 16$ , so there are 32 possible values of $D$ .

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