Worked solution: A rainbow-tree proof of Ringel's conjecture (2020)

The conjecture asks for an exact edge-decomposition: since a tree $T$ with $n$ edges has $n+1$ vertices and $K_{2n+1}$ has $n(2n+1)$ edges, the count $n \cdot (2n+1)$ matches exactly, so a decomposition must use precisely $2n+1$ copies of $T$, each edge of $K_{2n+1}$ covered once. Montgomery, Pokrovskiy and Sudakov (Geom. Funct. Anal. 2021, Theorem 1.2) prove this for every sufficiently large $n$, with no restriction on the shape or maximum degree of $T$.

The result sits on top of a long line of partial answers that the introduction of the paper surveys: Ringel's conjecture was previously known for special families (paths and stars classically; caterpillars, trees with at most $4$ leaves, and trees of bounded degree through work of Joos, Kim, Kühn and Osthus), while Ferber–Samotij and Adamaszek–Allen–Grosu–Hladký proved approximate versions covering almost all of $K_{2n+1}$ for trees of maximum degree $O(n/\log n)$. What remained open — and what makes the conjecture hard — is the union of extremes: trees that are simultaneously sparse, of arbitrary degree sequence, and structurally heterogeneous.

The proof does not decompose $K_{2n+1}$ directly. Instead it proves a stronger, seemingly unrelated statement — every edge-colouring of $K_{2n+1}$ by cyclic distance contains a rainbow copy of $T$ (next step) — and recovers the decomposition as a corollary, for all large $n$.