A 3×3 square is divided into 9 squares. A positive integer is inscribed in every square, such that the sums of the numbers in each row, column and diagonal are equal. Prove that this sum cannot be 2008.
Solution
Let m be the sum in each row, column and diagonal. Then we have the following situation:
x
y
m−x−y
a
b
m−a−b
m−x−a
m−b−y
x+a+b+y−m
Because the sum of the elements in the diagonals must be m we obtain: {x+b+x+a+b+y−m=mm−x−a+b+m−x−y=m which is equivalent to {2x+a+y+2b=2m2x+a+y−b=m If we subtract the equations we obtain 2b+b=2m−m which is equivalent to 3b=m. Hence 3∣m. But 3 is not a divisor for 2008, hence m=2008.
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Source: MathNet,
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