= Solution
The identities require $r=k$ and $k(k-1)=2\cdot3=6$, so $k=r=3$. Take treatments $1,2,3,4$ and the four blocks
$$
\boxed{\{1,2,3\},\quad\{1,2,4\},\quad\{1,3,4\},\quad\{2,3,4\}.}
$$
Each treatment is absent from exactly one block, hence occurs three times. Each pair is absent from the two blocks missing one of its members, hence occurs together in the other two. This is the required <symmetric balanced incomplete block design>; here $r-\lambda=1$ is indeed a square.
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