Problem 3. If $n$ is a natural number, is the number

$$
(2 n-3)(2 n-1)(2 n+1)(2 n+3)+16
$$

a perfect square? Is the number 2005$\cdot$2007$\cdot$2009$\cdot$2011 + 16 a perfect square?
Solution. Since

$$
(2 n-3)(2 n+3)=4 n^{2}-9 \text { and }(2 n-1)(2 n+1)=4 n^{2}-1
$$

we get

$$
\begin{aligned}
(2 n-3)(2 n-1)(2 n+1)(2 n+3)+16 & =\left(4 n^{2}-9\right)\left(4 n^{2}-1\right)+16 \\
& =16 n^{4}-40 n^{2}+25=\left(4 n^{2}-5\right)^{2}
\end{aligned}
$$

Thus, $(2 n-3)(2 n-1)(2 n+1)(2 n+3)+16$ is a perfect square.

The number 2005$\cdot$2007$\cdot$2009$\cdot$2011 + 16 is a perfect square, because the result from the first part of the problem can be applied for $n=1004$. According to this,

$$
2005 \cdot 2007 \cdot 2009 \cdot 2011+16=\left(4 \cdot 1004^{2}-5\right)^{2}=1008016^{2}
$$