`m n` squares of the equal size are arranged to form a rectangle of dmensin `mb yn ,w h e r ema n dn` are natural number. Two square will be called neighbours if they have exactly one common side. A number is written in each square such that the number written in any square is the arithmetic men of the numbers written in its neighboring squares. Show that this is possible only if all the numbers used are equal.

Explanation

Let mn squares of equal size are arranged to form a rectangle of dimensions m and n shown in the figure. The neighbiur of `x_(1)` are `x_(2),x_(3),x_(4),x_(5)` .
Similarly neighbours of `x_(5)` are `x_(6) , x_(1)` and `x_(7)` and that of `x_(7)` are `x_(5) , x_(4)` .
From the guven condition ,
`x_(1)=(x_(2)+x_(3)+x_(4)+x_(5))/(4)`
`implies 4x_(2)=x_(2)+x_(3)+x_(4)+x_(5)`
Also `X_(5)=(x_(1)+x_(6)+x_(7))/(3)`
`implies 3x_(5)=x_(1)+x_(6)+x_(7) and x_(7)=(x_(4)+x_(5))/(2)`
`implies 2x_(7)=x_(4)+x_(5)`

`therefore 4x_(1)=x_(2)+x_(3)+x_(4)+(x_(1)+x_(6)+x_(7))/(3)`
`implies 12x_(1)+3x_(2)=3x_(3)+3x_(4)=x_(1)=x_(6)+x_(7)`
`=3x_(2)+3x_(3)+3x_(4)+x_(1)+x_(6)+ (x_(4)+x_(5))/(2)`
`implies 24x_(1)=6x_(2)+6x_(3)+7x_(4)+2x_(1)+2x_(6)+x_(5)`
`implies 22x_(1)=6x_(2)+6x_(3)+7x_(4)+x_(5)+2x_(6)`
Where `x_(1),x_(2),x_(3),x_(4),x_(5),x_(6)` are all natural numbers and `x_(1)` is linearly expressed as the sum of `x_(2),x_(3),x_(4),x_(5) and x_(6)` where the sum of coefficients is equal to the coefficient of `x_(1)`.
i.e., 22=6+6+7+1+2
Which will happen only when
`x_(1)=x_(2)=x_(3)=x_(4)=x_(5)=x_(6)`
`implies ` all the numbers used are equal .