Problem Analysis #17

To solve the problem, we are given the function:

$$h(n) = n^2 + 12n - 12$$

and asked to determine how many positive integers $n$ exist such that $h(n)$ is a square number.

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Step 1: Expressing the Condition

We are looking for positive integers $n$ such that:

$$n^2 + 12n - 12 = k^2$$

for some nonnegative integer $k$. To simplify this, complete the square on the left-hand side:

$$n^2 + 12n - 12 = (n + 6)^2 - 48$$

So the equation becomes:

$$(n + 6)^2 - k^2 = 48$$

This is a difference of squares, which factors as:

$$(n + 6 - k)(n + 6 + k) = 48$$

Let:

• $u = n + 6 - k$
• $v = n + 6 + k$

Then:

• $uv = 48$
• $u < v$
• $v - u = 2k$ must be even ⇒ $u$ and $v$ must have the same parity

Also, since n + 6 $$\geq$$ 7 (as n $$\geq$$ 1), and k $$\geq$$ 0, both $u$ and $v$ must be positive integers.

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Step 2: Enumerating Valid Factor Pairs

List all positive factor pairs of 48 with $u < v$, and check which ones have the same parity:

| Pair | Same Parity? | Valid? | |------|---------------|--------| | (1, 48) | No | ❌ | | (2, 24) | Yes | ✅ | | (3, 16) | No | ❌ | | (4, 12) | Yes | ✅ | | (6, 8) | Yes | ✅ |

Only these three pairs satisfy the condition that $u$ and $v$ have the same parity.

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Step 3: Solving for $n$ in Each Case

For each valid pair $(u, v)$, compute:

$$n + 6 = \frac{u + v}{2}, \quad k = \frac{v - u}{2}$$

Case 1: $(u, v) = (2, 24)$

$$n + 6 = \frac{2 + 24}{2} = 13 \Rightarrow n = 7 \\ k = \frac{24 - 2}{2} = 11$$

Check: h(7) = 7^2 + 12 $$\cdot$$ 7 - 12 = 49 + 84 - 12 = 121 = 11^2 ✅

Case 2: $(u, v) = (4, 12)$

$$n + 6 = \frac{4 + 12}{2} = 8 \Rightarrow n = 2 \\ k = \frac{12 - 4}{2} = 4$$

Check: $h(2) = 4 + 24 - 12 = 16 = 4^2$ ✅

Case 3: $(u, v) = (6, 8)$

$$n + 6 = \frac{6 + 8}{2} = 7 \Rightarrow n = 1 \\ k = \frac{8 - 6}{2} = 1$$

Check: $h(1) = 1 + 12 - 12 = 1 = 1^2$ ✅

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Step 4: Verifying No Other Solutions

• We considered all positive factor pairs of 48.
• Only those where $u$ and $v$ have the same parity yield integer values for $k$.
• Negative factor pairs would lead to $n + 6 < 0$, contradicting the fact that n $$\geq$$ 1.
• Checking values of $n > 7$ manually shows no further perfect squares in $h(n)$.
• Also, solving the quadratic $n^2 + 12n - 12 = k^2$ for small $k$ confirms only the above three solutions.

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Final Answer

The positive integers $n$ for which $h(n)$ is a square number are:

$$n = 1,\ 2,\ 7$$

Thus, the total number of such positive integers is:

$$\boxed{3}$$