For n>10 (n ²+3n+1) ² >N >(n ²+3n) ² For -2>n (n ² +3n+2) ² >N>(n ² +3n+1) ² So for 10 >=n>=-2 the only variant is n=10

This is quite an easy one for a chinese olimpiad, isn't it so?

is there any simple solution please?

need help please

See there https://www.artofproblemsolving.com/community/c6h1285523

I edited the problem to more complicated one

$$n^4+6n^3+9n^2<N= n^4 + 6n^3 + 11n^2 +3n+31<n^4+6n^3+13n^2+12n+4$$$$\therefore (n^2+3n)^2<N<(n^2+3n+2)^2$$So $N$ must be equal to $(n^2+3n+1)^2$ $$\therefore N= n^4 + 6n^3 + 11n^2 +3n+31=n^4+6n^3+11n^2+6n+1$$$$\therefore 3n+31=6n+1$$$$\therefore n=\color{red}10$$ When $n=10$, $N=131^2$.

Yes, bounding always does kill problems like this.