Problem Analysis #17

To find the number of positive integers $n$ for which $h(n) = n^2 + 12n - 12$ is a square number, we set $h(n) = k^2$ for some nonnegative integer $k$.

This gives the equation:

We can complete the square on the left side:

Rearranging the terms, we get a difference of squares:

Let $a = n+6-k$ and $b = n+6+k$. Since $n \ge 1$ and $k \ge 0$, we have $b \ge 7$. Also, $a$ and $b$ must have the same parity because their sum is $2(n+6)$, which is even. Since their product is $48$ (an even number), both $a$ and $b$ must be even. Since $k \ge 0$, we must have $a \le b$.

The pairs of positive even integers $(a, b)$ that multiply to $48$ are:

1. $a = 2, b = 24$
2. $a = 4, b = 12$
3. $a = 6, b = 8$

For each pair, we can solve for $n$ using the formula $n = \frac{a+b}{2} - 6$:

• For $(2, 24)$: $n = \frac{2+24}{2} - 6 = 7$ (which gives $h(7) = 121 = 11^2$)
• For $(4, 12)$: $n = \frac{4+12}{2} - 6 = 2$ (which gives $h(2) = 16 = 4^2$)
• For $(6, 8)$: $n = \frac{6+8}{2} - 6 = 1$ (which gives $h(1) = 1 = 1^2$)

All three values of $n$ ($7, 2, 1$) are positive integers.

Thus, there are exactly $3$ positive integers $n$ for which $h(n)$ is a square number.

$\boxed{3}$