I'm trying to proof this for a magic trick.
Let $n\in\mathbb{N}$. There are real numbers $(A_{i,j,k})_{i,j,k=1,...,n}$ and a value $S$ so that $$\sum_{i=1}^nA_{\pi_1(i),\pi_2(i),\pi_3(i)}=S$$ for every permutation $\pi_1,\pi_2,\pi_3$. In that case you can choose numbers $a_i,b_j,c_k$ so that $A_{i,j,k}=a_i+b_j+c_k$ for every $i,j,k=1,...,n$ and $$S=\sum_{i=1}^na_i+\sum_{j=1}^nb_j+\sum_{k=1}^nc_k$$
I already proofed this for 2 dimensions(without c) by defining $a_1=x$,$b_j=A_{1,j}-x$. Then I defined $A_{1,j}=a_1+b_j$ and $a_i=A_{i,1}-b_1$ so that $A_{i,1}=a_i+b_1$. After that I went for the cases $i,j>1$. But in this case I didn't come that far. I thought about using $a_1:=x,b_2:=y,c_k:=A_{1,1,k}-x-y$. But then I don't know how to continue. How do I come to the point where I can say: $A_{i,j,k}=a_i+b_j+c_k$ if $i$ or $j$ or $k=1$. Could someone give me a hint on how to go on?