China Mathematical Competition

Obviously, every element of $A$ appears three times in all the ternary subsets. Then we have $$3(a_1+a_2+a_3+a_4)=(-1)+3+5+8=15,$$ or $a_1+a_2+a_3+a_4=5$. Therefore, the four elements of $A$ are $5-(-1)=6$, $5-3=2$, $5-5=0$, $5-8=-3$, respectively. The answer is $A=\{-3,0,2,6\}$. $\square$