An archipelago consists of $n$ islands. Some of the islands are pairwise connected by bridges (maybe zero or more than one). It is known that from any island one can arrive to any other island, and that the first island has $1$ bridge, the second $4$, ..., the $n$-th has $n^2$ bridges. Find all $n$ for which such an archipelago can exist.
(Belarussian math contest 1992)
A necessary condition is that the number of island-bridge incidences (or bridge-ends) is even (as it is twice the total number of bridges). As $1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$, this means that $n\equiv 0\pmod 4$ or $n\equiv 3\pmod 4$. Also, it is necessary that $n^2\le 1^2+2^2+\ldots+(n-1)^2$ or else we're forced to have some bridge on island $n$ loop back to the same island. This condition is equivalent to $n^2\le \frac{(n-1)n(2n-1)}6$, i.e., $6n\le (n-1)(2n-1)$, or: $n\ge 5$.
These conditions are also sufficient: Suppose that $n\ge 5$ and $n\equiv 0$ or $\equiv 3\pmod 4$. Then we can assign $1^2+2^2+\ldots+n^2$ bridge-ends to the islands such that the $i$th island obtains $i^2$ bridge-ends. Our task is to match bridge-ends in pairs to form bridges. It is straightforward to first join island $i$ with $i+1$ in order to ensure that we have connectivity as required (this uses one bridge-end for $i=1$ and for $i=n$, and two bridge-ends for all other $i$). Among all pairings among the remaining bridge-ends, pick one that minimizes the number $\ell$ of loops (i.e., bridges running from an island to itself). Assume $\ell>0$. Let island $a$ have a loop between bridge-ends $u$ and $v$. As the remaining at most $n^2-2$ bridge-ends on island $a$ cannot use up the at least $1^2+2^2+\ldots+(n-1)^2$ bridge-ends on the other islands, there exists a bridge between an end $x$ on an island $b\ne a$ and an end $y$ on island $c\ne a$ (whereas we do allow $b=c$). Then we remove bridges $uv$ and $xy$ and replace them with $ux$ and $vy$. Note that this does not destroy connectivity (we can walk from $b$ to $c$ via $a$) and we end up with at most $\ell-1$ loops, contradicting minimality of $\ell$. We conclude that either $\ell=0$, as desired.
I found another proof for sufficiently with induction. Since the configurations for a base cases $n=7$ and $n=8$ are easy to find I'll skip to induction step $n \to n+4$:
Say we have $n+4$ islands. Then connect the island $A_{n+3}$ with $(n+3)^2$ bridges with $A_{n+4}$ and this one with another $2n+3$ bridges to $A_{n+2}$ and this one with another $(n+1)^2$ bridges to $A_{n+1}$. Now we have to make $4$ more bridges from $A_{n+4}$.
By induction hypothesis we can arrange bridges between island $A_1,A_2,...A_n$. It is not difficult to see that $A_n$ is connected with some of the island $A_1,A_2,...A_{n-1}$ with at least $3$ bridges (suppose it isn't true, then $n^2\leq 1+2(n-2) \Longrightarrow (n-1)^2\leq -2)$. Call this island $B$. Then we can destroy $2$ bridges between $A_n$ and $B$. Then connect $A_{n+4}$ with $A_n$ with $2$ new bridge and the same with $B$.