Problem Analysis #17

To determine the number of positive integers $n$ for which $h(n) = n^2 + 12n - 12$ is a square number, we start by setting $h(n) = k^2$ for some integer $k$. This gives us the equation:

Rearranging terms, we get:

This is a quadratic equation in $n$. For $n$ to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant $D$ is given by:

For $D$ to be a perfect square, let $D = m^2$ for some integer $m$. Then:

Rearranging gives:

This can be factored as a difference of squares:

We need to find pairs of factors of 192 that differ by an even number (since $m - 2k$ and $m + 2k$ are both even). The factor pairs of 192 are:

We check each pair to see if they satisfy the condition $m - 2k$ and $m + 2k$ being even and leading to valid $k$ and $n$:

1. $(1, 192)$: $m = 97.5$ (not an integer).
2. $(2, 96)$: $m = 49$, $k = 12$. Then $n = \frac{-12 \pm 49}{2}$. This gives $n = 18.5$ (not an integer) or $n = -31.5$ (not positive).
3. $(3, 64)$: $m = 33.5$ (not an integer).
4. $(4, 48)$: $m = 26$, $k = 10$. Then $n = \frac{-12 \pm 26}{2}$. This gives $n = 7$ (positive integer) or $n = -19$ (not positive).
5. $(6, 32)$: $m = 19$, $k = 8$. Then $n = \frac{-12 \pm 19}{2}$. This gives $n = 3.5$ (not an integer) or $n = -17$ (not positive).
6. $(8, 24)$: $m = 16$, $k = 6$. Then $n = \frac{-12 \pm 16}{2}$. This gives $n = 2$ (positive integer) or $n = -14$ (not positive).
7. $(12, 16)$: $m = 14$, $k = 4$. Then $n = \frac{-12 \pm 14}{2}$. This gives $n = 1$ (positive integer) or $n = -11$ (not positive).

The valid positive integer solutions for $n$ are $n = 1, 2, 7$. Therefore, the number of positive integers $n$ for which $h(n)$ is a square number is: