Question

The entries of an $n\times n$ array of numbers are denoted by $a_{ij},1\le i,j\le n.$ The sum of any n entries situated on different rows and different columns is the same. Prove that there exist numbers $x_1,x_2,...x_n$ and $y_1,y_2,...y_n,$ such that $a_{ij}=x_i+x_1+y_j,$ for all i, j.

• A. $a_{ij}=x_i+x_1+y_j,$ for all i, j, as desired.
• B. $a_{ij}=x_i-x_1+y_j,$ for all i, j, as desired.
• C. $a_{ij}=x_i+x_1-y_j,$ for all i, j, as desired.
  
• D. $a_{ij}=x_i-x_1-y_j,$ for all i, j, as desired.

Answer 1

The correct option is A $a_{ij}=x_i+x_1+y_j,$ for all i, j, as desired.

Consider n entries situated on different rows and different columns $a_{iji},i=1,2,...,n.$ Fix $k$ and $l,1\le k<l\le n$ and replace $a_{kjk}$ and $a_{ljt}$ with $a_{kjl}$ and $a_{ljk}$
respectively. It is not difficult to see the new n entries are still situated on different rows and different columns. Because the sums of the two sets of n entries are equal, it follows that
$a_{kjk}+a_{ljl}=a_{kjl}+a_{ljk}.$
Now, denote $x_1,x_2,...,x_n$
the entries in the first column of the array and by $x_1,x_1+y_2,x_1+y_3,....,x_1+y_n$ the entries in the first row (in fact, we have defined $x_k=a_{k1},$
for all $k,y_0=0$ and
$yk=a_{1k}-a_{k1}$
for all $k\ge 2$
The equality $a_{ij}=x_i+y_j,$
stands for all $i,j$ with $i=1$ or $j=1$.
Now, consider $i,j>1$. From (*)
we deduce $a_{11}+a_{ij}=a_{1j}+a_{il}$
hence $x_1+a_{ij}=x_i+x_1+y_j$
or, $a_{ij}=x_i+x_1+y_j,$
for all $i,j,$ as desired.

$x_1$	$x_1+y_2$	...	$x_1+y_j$	...	$x_1+y_n$
$x_2$
. . .
$x_i$			$a_{ij}$
. . .
$x_n$