smc

The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z)\cdot Q(z)$ has degree $2+3=5,$ we conclude that $R(z)-P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant. It now suffices to illustrate an example for which $N=1$: Take $$\begin{array}{rl}P(z)& =z^2+1,\\ Q(z)& =z^3+2,\\ R(z)& =(z+1{)}^6+P(z)\cdot Q(z).\end{array}$$ Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions. We need to find the solutions to $$\begin{array}{rl}P(z)\cdot Q(z)& =(z+1{)}^6+P(z)\cdot Q(z)\\ 0& =(z+1{)}^6.\end{array}$$ Clearly, the only distinct complex root is $-1,$ so our answer is $N=\boxed{1}.$