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 3 of 4: Eliminate the first two candidate vertex sets using modulo 4

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In plain words

Checking squares modulo $4$ is a fast filter: no sum of two integer squares can ever leave remainder $3 \pmod{4}$ , which rules out $47$ and $67$ without any trial and error.

\{a,b,c\}=\{3,6,9\} \implies t=137,\ x^2+y^2=47 \equiv 3 \pmod{4}; \qquad \{a,b,c\}=\{1,4,7\} \implies t=117,\ x^2+y^2=67 \equiv 3 \pmod{4}

Detailed analysis

If $\{a, b, c\} = \{3, 6, 9\}$ , then $3t = 3^2 + 6^2 + 9^2 + 285 = 411$ , so $t = 137$ . On the side connecting vertices $3$ and $9$ , the two interior numbers $x, y$ must satisfy $x^2 + y^2 = 137 - (3^2 + 9^2) = 47 \equiv 3 \pmod{4}$ , which is impossible because $x^2, y^2 \equiv 0$ or $1 \pmod{4}$ implies $x^2 + y^2 \equiv 0, 1, 2 \pmod{4}$ . Similarly, if $\{a, b, c\} = \{1, 4, 7\}$ , then $3t = 1^2 + 4^2 + 7^2 + 285 = 351$ , so $t = 117$ ; on the side between $1$ and $7$ , the interior numbers must satisfy $x^2 + y^2 = 117 - (1^2 + 7^2) = 67 \equiv 3 \pmod{4}$ , again impossible.