Let an $11\times 11$ square filled with numbers from 1 to 121. Is there a construction such that every $3 \times 3$ square has the same sum? Is there any generalization of the problem for $6n-1 \times 6n-1 $ squares?
My work: For the case of $5 \times 5$ square I found that we can achieve it (every $3 \times 3$ subsquare has sum 93)