Problem： n

^{2}+ 6n + 24 is a perfect square number. Find integer n.问题：是完全平方数。求 整数n.

The problem seems clueless. It says that is a perfect square number, i.e. the square of an integer, but we do not know which integer.

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But we know perfect square numbers are rare, and for integer n the expression will leave big gaps even if we increase / decrease n by one only. Chances are, there are only a few solutions.}

In algebra, we will talk about that an expression is a perfect square, e.g.,

n^{2} + 2 n + 1 = (n+1)^{2}, and (n^{2} + n + 1/4) = (n+ 1/2)^{2}.

Will an expression in perfect square, for certain n value, generate a perfect-square number? And if an expression, for some n value, generates perfect square numbers, then is the expression also a perfect square (as in algebra)? The answer to both questions are not necessarily true.

However, there is one important relation, for example, if n is an integer, then n+1 must also be an integer, and (n+1)^{2} is the square of integer n+1, therefore:

n is an integer –> is a perfect square

This analysis leads to a solution idea. Actually,

Now what? For n integer, we have two perfect squares whose difference is 15. The two perfect squares are , and resp. .

Can you find two perfect numbers whose difference is 15? Yes for sure!

Try it yourself! If you want to find (the answer or more hints), check it out!

Solution(s) that does not base on completing-the-square is possible. We just need other clues.