When is n⁴ + n³ + n² + n + 1 a perfect square?

Answer. $n=3$ only ($N=121=11^2$).

Idea. Trap $4N$ strictly between two consecutive squares of the same parity, leaving a single possible intermediate square, then solve for $n$.

Bounds. We have
$$(2n^2+n)^2=4n^4+4n^3+n^2<4N=4n^4+4n^3+4n^2+4n+4,$$
and
$$(2n^2+n+2)^2=4n^4+4n^3+9n^2+4n+4,$$
so $(2n^2+n+2)^2-4N=5n^2>0$. Hence
$$(2n^2+n)^2<4N<(2n^2+n+2)^2.$$

Parity. If $N=m^2$ then $4N=(2m)^2$ is an even square strictly between the squares of $2n^2+n$ and $2n^2+n+2$, so $2m=2n^2+n+1$. Squaring,
$$4N=(2n^2+n+1)^2=4n^4+4n^3+5n^2+2n+1,$$
and comparing with $4N=4n^4+4n^3+4n^2+4n+4$ gives $n^2-2n-3=0$, i.e. $(n-3)(n+1)=0$.

Conclusion. $n=3$.

Check. $81+27+9+3+1=121$ ✓. (Compare: $n=1$ gives $5$, $n=2$ gives $31$, $n=4$ gives $341$.)

Lesson. To solve "polynomial = square", sandwich the polynomial (suitably scaled) between two consecutive squares and treat the few possible intermediate values.