Problem： Number n

^{2}+ 6n + 24 is a perfect square number. Find integer n.

Last time, we have given an introductory analysis. Now we will continue to a full solution.

**Analysis and Solution:**

__Methods based on completing the square).__

For integer n, (n+3)^{2}is a perfect square, and

according to the statement in question, n^{2} +6n + 24 = (n+3)^{2} + 15 is also a perfect square.

Therefore, we have two perfect squares whose difference is 15.

The number n must have a relatively small (absolute) value, since the perfect squares of different large numbers will have a bigger difference – we guess n has only one digit.

*Approach 1. *

Looking for two perfect squares whose difference is 15.

List the perfect squares of all one-digit numbers:

n 0 1 2 3 4 5 6 7 8 9

0 1 4 9 16 25 36 49 64 81

By some trial and error, we find that: 15 = 4^{2} – 1^{2} = 8^{2} – 7^{2}.

Pick out the smaller of the pair of perfect square numbers (in 4^{2} – 1^{2}, pick out 1; in 8^{2} – 7^{2}, pick out 7), then match with the smaller perfect number to find solutions (do not forget (n+3) can be negative! There are four matches: n+3 = -1 or 1 or -7 or 7.)

*Approach 2. *

Let K = n+3. Then we have K^{2} + 15 = L^{2}, where K, L are both integers.

Now rewrite to: 15 = L^{2} – K^{2}

and using L^{2} – K^{2} = (L+K) (L – K)

So 15 = (L+K)(L-K)

On the other hand, 15 = 5 × 3 = 15 × 1.

Find integer K – focus on positive integer solution, then negative solution.

Once found K, use n = K-3 to get the value of n. // Our readers, try provide the rest steps.

*Approach 3. (Not based on “completing-the-square”)*

n

^{2}+ 6n + 24 = (n+4)(n+6) – 4n,

We guess that n is an even number (can you justify it?) so let n = 2m, we have:

n^{2} + 6n + 24 = (2 m+4)(2 m+6) – 4(2m)

Factor out constant 4, we shall have another perfect square number,

(m+2) (m+3) – 2m

Now let k = m+2, we have

k (k+1) – 2(k-2) = 4 + k(k-1)

Now (i) 4 is a perfect square, therefore we can let k (k-1) = 0 which leads to k = 0, k = 1.

(ii) 16 = 4 + 12 is a perfect square, therefore we can let k (k-1) = 12 which leads to k = 4 or k = -3.

To find n , just note that n = 2m and k = m+2, so n = 2 (k-2).

// Our readers, try provide the rest steps.

Finally, we give:

- Reference answer: whatever the approach, you can get 4 answers: they are n = -4 or -2 or -10 or 4.
- Question for thought: Can you justify that the solution n in this question must be an even number?

We have applied this in 3rd approach of our solution. Please try justify.