$\left[\begin{array}{l}\text { Factorization } \\ \text { [Evenness and Oddness] }\end{array}\right]$

Can the difference of two numbers of the form $n^{2}+4 n$ (where $n$ is a natural number) equal $1998 ?$
$n^{2}+4 n-m^{2}-4 m=(n+m+4)(n-m)$. Note that both factors have the same parity. If both factors are odd, then their product is odd. If the factors are even, then their product is divisible by 4; but 1998 is not divisible by 4.

Answer

It cannot.