Erdős Problem 794: a nine-vertex hypergraph counterexample
Statement and claim.
The assertion that every $3$-uniform hypergraph on $3n$ vertices with at least $n^3+1$ edges contains either four vertices with three edges or five vertices with seven edges is false. We verify the counterexample attributed to Phillip Harris at
https://www.erdosproblems.com/794. All edges and induced edge counts are checked directly below.
Construction.
Set $n=3$ and partition nine vertices into \[
A=\{1,2,3\},\quad B=\{4,5,6\},\quad C=\{7,8,9\}.
\] Take every triple with one vertex in each part, and add the triple $A$ itself. There are $3^3+1=28$ edges.
No four vertices span three edges.
For a four-element set $S$, put $a=|S\cap A|$, $b=|S\cap B|$, $c=|S\cap C|$. Its edge count is \[
abc+{\bf1}_{a=3},\qquad a+b+c=4.
\] If all three counts are positive, their multiset is $\{2,1,1\}$; there are exactly two transversal edges, and $a\ne3$. If one count is zero, there are no transversal edges and at most the single added edge. Thus every four vertices span at most two edges.
The second forbidden configuration is redundant.
In any $3$-uniform hypergraph on five vertices, each edge belongs to exactly two of the five four-vertex subsets. If there are seven edges, the sum of those five induced edge counts is $14$, and some count is at least three. Consequently the absence of the first configuration already excludes the second. For this particular example there is an even stronger check: on five vertices the count is $abc+{\bf1}_{a=3}$. With all counts positive, the patterns are $(3,1,1)$ or $(2,2,1)$, up to permutation, and the count is at most four. With a zero count it is at most one.
Scope.
The explicitly stated finite assertion is fully disproved. This does not claim the sharp asymptotic Turán density for the four-vertex, three-edge configuration; that is a different extremal problem.
Completion estimate.
COMPLETION: 100%