For how many integers n is (n)/(20 - n) the square of an integer?

2002 AMC 10B Problem 16 — Solution

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16.

For how many integers $n$ is $\dfrac{n}{20 - n}$ the square of an integer?

$1$

$2$

$3$

$4$

$10$

Suppose $\dfrac{n}{20 - n} = k^2$ for some integer $k \ge 0.$ Solving, $n = \dfrac{20k^2}{k^2 + 1}.$

Since $k^2$ and $k^2 + 1$ share no common factor, $k^2 + 1$ must divide $20.$ This happens only for $k = 0, 1, 2, 3,$ giving $k^2 + 1 = 1, 2, 5, 10.$

The corresponding values $n = 0, 10, 16, 18$ are all integers, so there are $4$ such $n.$

Thus, the correct answer is D.