Quite correct.

In fact, if we assume the product to be a perfect square, any prime > 3 that is a factor of any of the five integers must be a factor of only one of the integers and must have an even power, as your example demonstrates.

So each of the integers, in its prime factorization, can have only 2 or 3 with an odd power, and so each is one of the following forms:

Since there are five integers, two of the integers have one of these forms.

Let's go case by case with these forms.

The only squares close enough together to fit within five consecutive integers are 1 and 4, but

is not a perfect square. This eliminates the first form as a possibilty.

If two of the integers have the form

, then their difference being 4 or less impies that there are two squares with a difference of 2 or less. But no two squares exist with a difference of 1 or 2. This eliminates the second form as a possibilty.

Similarly, if two of the integers have the form

, then their difference of less than 5 implies there are two squares with a difference of 1, which again is impossible. This eliminates the third form as a possibilty.

Of course, there cannot be two multiples of 6 within any five consecutive integers, so the fourth form is also impossible.

Since all the possible forms are eliminated, the product of five consecutive integers cannot be a perfect square.