Problem 2

Given a triangular arrangement of

9

circles along the perimeter of a triangle, with

4

circles on each of the three sides (three vertex circles shared by adjacent sides and two interior circles on each side), each of the numbers

1, 2, \ldots, 9

is to be written into one of these circles, so that each circle contains exactly one of these numbers and: (i) the sums of the four numbers on each side of the triangle are equal; (ii) the sums of the squares of the four numbers on each side of the triangle are equal. Find all ways in which this can be done.

Step 2 of 4: Use modulo 3 arithmetic to narrow the vertices to three residue classes

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In plain words

Once the vertices are forced to share the same remainder modulo $3$ , they must exhaust one of the three residue classes $\{1, 4, 7\}$ , $\{2, 5, 8\}$ , or $\{3, 6, 9\}$ .

a + b + c \equiv 0 \pmod{3}, \quad a^2 + b^2 + c^2 \equiv 0 \pmod{3} \implies \{a, b, c\} \in \{\{3, 6, 9\}, \{1, 4, 7\}, \{2, 5, 8\}\}

Detailed analysis

Every integer square satisfies $x^2 \equiv 0$ or $1 \pmod{3}$ , so $a^2 + b^2 + c^2 \equiv 0 \pmod{3}$ holds if and only if either all three of $a, b, c$ are divisible by $3$ or none of them is divisible by $3$ . Combining this with $a + b + c \equiv 0 \pmod{3}$ shows that all three vertices must be congruent modulo $3$ . Since there are only three numbers in each residue class, $\{a, b, c\} \in \{\{3, 6, 9\}, \{1, 4, 7\}, \{2, 5, 8\}\}$ .