When is it a square?

If $F(n) = n^4 + n^3 + n^2 + n + 1$ then $F(1) = 5$ and $F(2) = 31$ , so for $F(n)$ to be a square we need $n \ge 3$ .

Suppose now that $n \ge 3$ and that $F(n) = m^2$ for some positive integer $m$ . Now $(n^2)^2 = n^4 < F(n) < n^4 + 2n^3 + n^2= (n^2+n)^2$ , so we deduce that $n^2 < m < n^2 + n$ . Thus $m = n^2 + b$ for some $1 \le b \le n-1$ . Thus $n^4 + n^3 + n^2 + n + 1 \; = \; F(n) \; = \; (n^2 + b)^2 \; = \; n^4 + 2bn^2 + b^2$ Since $1 \le b \le n-1$ , we deduce that $b^2= n+1$ , and hence $2b = n+1$ as well. Thus $b^2 = 2b$ , and hence we must have $b=2$ , $n=3$ .

The only solution is $n = \boxed{3}$ .