Problem 4

Consider a

2018\times2019

board with an integer written in each unit square. Two unit squares are called neighbours if they share a common edge. In each turn, some unit squares are chosen; then, for each chosen square, the average of all its neighbours is computed, and finally, after all these computations are done, the number in each chosen square is replaced by its corresponding average. Is it always possible to make the numbers in all squares equal after finitely many turns?

Step 3 of 6: Reflecting an equilibrium board preserves equilibrium

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\text{degree}\ 2\to3:\ 2(a+b+c)\equiv a;\quad \text{degree}\ 3\to4:\ 4(a+b+c+d)\equiv a;\quad \text{degree}\ 2\to4:\ 4(2a+b+c)\equiv a

Detailed analysis

Reflect an equilibrium board across one of its edges to double its size in that direction; the reflected copy repeats the same values, so the two seam columns (or rows) become equal. A cell already having $4$ neighbours keeps the same neighbours after reflection, so its equilibrium is untouched. A boundary cell with $2$ neighbours $b,c$ (so $a\equiv3(b+c)$ , i.e. $b+c\equiv2a$ ) that gains one more neighbour equal to itself now satisfies the displayed $2(a+b+c)=2(a+2a)=6a\equiv a$ identity; if it gains two such neighbours (a corner reflected in two directions) the displayed $4(2a+b+c)=4(2a+2a)=16a\equiv a$ identity holds. A cell with $3$ neighbours $b,c,d$ (so $a\equiv2(b+c+d)$ , i.e. $b+c+d\equiv3a$ ) that gains one more neighbour equal to itself satisfies $4(a+b+c+d)=4(a+3a)=16a\equiv a$ . In every case the reflected board is again in equilibrium modulo $5$ .