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Algebra Difficulty 5.4 AIME, harder Prove it North Macedonia

A 3×33 \times 3 square is divided into 99 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 20082008.

Solution

Let mm be the sum in each row, column and diagonal. Then we have the following situation:

xxyym−x−ym-x-y
aabbm−a−bm-a-b
m−x−am-x-am−b−ym-b-yx+a+b+y−mx+a+b+y-m

Because the sum of the elements in the diagonals must be mm we obtain:
{x+b+x+a+b+y−m=mm−x−a+b+m−x−y=m which is equivalent to \begin{cases} x+b+x+a+b+y-m = m \\ m-x-a+b+m-x-y = m \end{cases} \text{ which is equivalent to}
{2x+a+y+2b=2m2x+a+y−b=m \begin{cases} 2x+a+y+2b = 2m \\ 2x+a+y-b = m \end{cases}
If we subtract the equations we obtain 2b+b=2m−m2b + b = 2m - m which is equivalent to 3b=m3b = m. Hence 3∣m3 \mid m. But 33 is not a divisor for 20082008, hence m≠2008m \neq 2008.

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