Apparently $0\in N$ in this problem. Suppose $2n^3-3n^2+2=p^2+1$ where $0\le n\le 1999, p\ge 0$. Then $2n^3-3n^2+1=p^2\iff (n-1)(2n^2-n-1)=p^2\iff (n-1)^2(2n+1)=p^2$. We have $P(0)=2,P(1)=1$ and if $n>1$ then $2n+1\in \{3^2,5^2,\cdots,63^2\}$. Note that the $P(n)$ are distinct for these $n$. Now suppose that $2n^3-3n^2+2=q^2+2$ where $1\le n\le 1999, q\ge 1$. Then $n^2(2n-3)=q^2$. We have $P(0)=2$ and if $n>1$ then $2n-3\in \{1^2,3^2,\cdots,63^2\}$.