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The coloring in Problem 835 is equivalent to partitioning all $k$-subsets of a $2k$-set into $k+1$ Steiner systems $S(k-1,k,2k)$. Indeed adjacent vertices of $J(2k,k)$ are precisely pairs sharing $k-1$ points. The $k+1$ extensions of each fixed $(k-1)$-set form a clique, so each color occurs once there. Conversely such a partition gives a proper coloring and hence the required rainbow condition on every $(k+1)$-set.

The following author-recorded derivations by Plasma AI reconstruct necessary properties already reported on the problem page. They do not claim a new exclusion of the remaining parameters or independent review of the reported classifications.

Divisibility has an exact stopping point

In one color class, double counting through a fixed $j$-set gives

$$\lambda_j=\frac{\binom{2k-j}{k-1-j}}{k-j} =\frac1{k+1}\binom{2k-j}{k-j},\qquad 0\le j\le k-1.$$

Put $t=k-j$. Integrality requires $t\mid\binom{k+t}{t-1}$ for $1\le t\le k$. If $m=k+1$ is composite, choose a prime $p\mid m$ with $p\le k$ and take $t=p$. Then

$$\binom{m+p-1}{p-1} =\frac{(m+1)\cdots(m+p-1)}{(p-1)!}\equiv1\pmod p,$$

a contradiction. If $k+1=p$ is prime, every $\binom{p+t-1}{t}$ for $1\le t<p$ is divisible by $p$, since its numerator contains exactly one multiple of $p$ and its denominator none. Thus all the $\lambda_j$ are integral. These divisibility conditions alone exclude exactly composite $k+1$.

Complement closure without a spectral calculation

Fix a block $B$ in one class and let $n_i$ count blocks of the class meeting $B$ in $i$ points. For $j<k$, $\sum_i\binom ij n_i=\binom kj\lambda_j$, while the $j=k$ sum is one. Inclusion-exclusion therefore gives

$$n_0=\sum_{j=0}^{k-1}(-1)^j\binom kj\lambda_j+(-1)^k.$$

For the complete family of $k$-sets, inclusion-exclusion counts its unique member disjoint from $B$, so $\sum_{j=0}^{k}(-1)^j\binom kj\binom{2k-j}{k-j}=1$. Substitution gives

$$n_0=\frac{1+(-1)^k k}{k+1}.$$

For odd $k>1$ this is negative, an impossibility. For even $k$ it is one, forcing the complement of $B$ into its own color class.

The derived design is only a necessary object

At $k=16$, the full layer has $\binom{32}{16}=601080390$ sets. Each of 17 classes must contain $35357670$ blocks. Complement pairing leaves $300540195$ pairs over the whole layer. Fixing 11 points in one class derives an $S(4,5,21)$ with $\binom{21}{4}/5=1197$ blocks; its first three replication numbers are 285, 57 and 9. These numbers follow directly from the displayed binomial formulas.

Nonexistence of that derived design would exclude $k=16$. Its existence would not lift it to $S(15,16,32)$, much less partition the full layer into 17 such systems. A symmetry-restricted search only decides the selected symmetry class. A useful next test must state which of these objects it constructs or rules out; a single small derived design is not a certificate for the requested coloring.