We solve the currently smallest open case in the 1976 problem of Molluzzo on $\mathbb{Z}/m\mathbb{Z}$, namely the case $m=4$. This amounts to constructing, for all positive integer $n$ congruent to $0$ or $7 \bmod{8}$, a sequence of integers modulo $4$ of length $n$ generating, by Pascal's rule, a Steinhaus triangle containing $0,1,2,3$ with equal multiplicities.