MOPSS, 21st March, 2026

Denote the points by \(A, B, C, D\) from left to right. Let \(x, y, z\) be the distances between consecutive points, that is, \(x = AB, y = BC, z = CD\). For \(P\) in \({A, B, C, D}\), let \(S_P\) be the sum of the distances from \(P\) to the other three points. Then, we have

\(\begin{align*} S_A & = AB + AC + AD \\ & = x + (x+y) + (x+y+z) \\ & = 3x + 2y + z, \\ S_B & = AB + BC + BD \\ & = x + y + (y+z) \\ & = x + 2y + z, \\ S_C & = AC + BC + CD \\ & = (x+y) + y + z \\ & = x + 2y + z, \\ S_D & = AD + BD + CD \\ & = (x+y+z) + (y+z) + z \\ & = x + 2y + 3z. \end{align*} \\)

Note that \(S_B = S_C\). Thus, at least two of the numbers \(S_A, S_B, S_C, S_D\) are equal. This is only the case for the first set of numbers, \(29, 29, 35, 37\). Also observe that for \(x = 3, y = 11, z = 4\), we have \(S_A = 35, S_B = S_C = 29, S_D = 37\). This shows that only the first set of numbers can occur as the sums of the distances from the four points to the other three points in some order.

Let us first prove the following claim.

Claim. The parity of the number of multiples of \(3\) on the blackboard is an invariant.

Proof of the claim. Let \(a\) and \(b\) be the two numbers chosen in an operation. If \(a\) and \(b\) are both multiples of \(3\), then \(a^2 + b^2 + 2\) is not divisible by \(3\), and hence, \(\gcd(a^2 + b^2 + 2, a^2b^2 + 3)\) is not divisible by \(3\). If exactly one of \(a\) and \(b\) is a multiple of \(3\), then \(a^2 + b^2 + 2\) and \(a^2b^2 + 3\) are both divisible by \(3\), and hence, \(\gcd(a^2 + b^2 + 2, a^2b^2 + 3)\) is divisible by \(3\). If neither \(a\) nor \(b\) is a multiple of \(3\), then \(a^2 + b^2 + 2\) is not divisible by \(3\), and hence, \(\gcd(a^2 + b^2 + 2, a^2b^2 + 3)\) is not divisible by \(3\). Therefore, the parity of the number of multiples of \(3\) on the blackboard remains unchanged after each operation.

Initially, there are \(33\) multiples of \(3\) on the blackboard, so the parity of the number of multiples of \(3\) is odd. After performing \(99\) operations, there is only one number left on the blackboard, and it must be a multiple of \(3\). Let \(x, y\) be the last two numbers chosen in the final operation. Then, the last number on the blackboard is \(\gcd(x^2 + y^2 + 2, x^2y^2 + 3)\). This shows that \(3\) divides \(x^2y^2 + 3\), which implies that \(3\) divides \(xy\). Therefore, \(9\) does not divide \(x^2y^2 + 3\), and hence, \(9\) does not divide \(\gcd(x^2 + y^2 + 2, x^2y^2 + 3)\). This shows that the last number on the blackboard is divisible by \(3\), but is not divisible by \(9\), and hence is not a perfect square.