Macedonian Mathematical Competitions

Because $(2n-3)(2n+3) = 4n^2 - 9$ and $(2n-1)(2n+1) = 4n^2 - 1$, we have
$$(2n-3)(2n-1)(2n+1)(2n+3) + 16 = (4n^2 - 9)(4n^2 - 1) + 16.$$
Expanding:
$$(4n^2 - 9)(4n^2 - 1) = 16n^4 - 4n^2 - 36n^2 + 9 = 16n^4 - 40n^2 + 9.$$
So,
$$(2n-3)(2n-1)(2n+1)(2n+3) + 16 = 16n^4 - 40n^2 + 9 + 16 = 16n^4 - 40n^2 + 25.$$
Notice that
$$16n^4 - 40n^2 + 25 = (4n^2 - 5)^2.$$
Therefore, $(2n-3)(2n-1)(2n+1)(2n+3) + 16$ is the square of a positive integer for every positive integer $n$.

For the number $2005 \times 2007 \times 2009 \times 2011 + 16$, if we set $n = 1004$, then
$$2005 \times 2007 \times 2009 \times 2011 + 16 = (4 \times 1004^2 - 5)^2 = 1008016^2.$$
Thus, $2005 \times 2007 \times 2009 \times 2011 + 16$ is also the square of a positive integer.