New Year's Countdown Day 15: Anti-magical Diagonals

Algebra Level 5

Define a heterosquare to be a square grid containing consecutive positive integers starting from 1 such that the sums of the integers in each row, column, and long diagonal are all different. Furthermore, define an anti-magic square to be a heterosquare whose sums form a consecutive sequence of integers.

If there exists a 15 × 15 15 \times 15 anti-magic square, then there are n n possible values d 1 , d 2 , … , d n d_1, d_2, \dots, d_n for the sum of its two diagonal sums. Find the value of n ∑ i = 1 n d i . n \displaystyle \sum_{i = 1}^n d_i.


Note: A diagonal sum is the sum of the numbers along a long diagonal of the square. For example, the two diagonal sums of the following square are 1 + 5 + 9 = 15 1 + 5 + 9 = 15 and 3 + 5 + 7 = 15 : 3 + 5 + 7 = 15: 1 2 3 4 5 6 7 8 9 \begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 4 & 5 & 6 \\ \hline 7 & 8 & 9 \\ \hline \end{array}


The answer is 13560.

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Dec 15, 2017

Let's derive a general result for a k × k k \times k anti-magic square, where k k is any positive integer greater than 2.

First, we use this fact: if a k × k k \times k anti-magic square exists, then there are only two possible sequences that can represent the sums of its rows, columns, and diagonals: k 3 − k − 2 2 , k 3 − k 2 , … , k 3 + 3 k 2 \frac{k^3 - k - 2}{2}, \frac{k^3 - k}{2}, \dots, \frac{k^3 + 3k}{2} and k 3 − k 2 , k 3 − k + 2 2 , … , k 3 + 3 k + 2 2 . \frac{k^3 - k}{2}, \frac{k^3 - k + 2}{2}, \dots, \frac{k^3 + 3k + 2}{2}. (See this problem or the discussion on the open problem page for a proof of why this is the case.) Call these sequences anti-magical sequences . The sum of the numbers in each anti-magical sequence is

1 2 ( 2 k + 2 ) ( k 3 − k − 2 2 + k 3 + 3 k 2 ) = ( k + 1 ) ( k 3 + k − 1 ) = k 4 + k 3 + k 2 − 1 1 2 ( 2 k + 2 ) ( k 3 − k 2 + k 3 + 3 k + 2 2 ) = ( k + 1 ) ( k 3 + k + 1 ) = k 4 + k 3 + k 2 + 2 k + 1. \begin{aligned} \dfrac{1}{2}(2k + 2) \left ( \dfrac{k^3 - k - 2}{2} + \dfrac{k^3 + 3k}{2} \right ) &= (k + 1)(k^3 + k - 1) = k^4 + k^3 + k^2 - 1 \\ \dfrac{1}{2}(2k + 2) \left ( \dfrac{k^3 - k}{2} + \dfrac{k^3 + 3k + 2}{2} \right ) &= (k + 1)(k^3 + k + 1) = k^4 + k^3 + k^2 + 2k + 1. \end{aligned}

The sum of all the integers from 1 to k 2 k^2 inclusive is equal to k 2 ( k 2 + 1 ) 2 = k 4 + k 2 2 . \frac{k^2(k^2 + 1)}{2} = \frac{k^4 + k^2}{2}. Note that this is equal to both the sum of all the columns and the sum of all the rows, because both cases cover all the numbers on the square. Also note that the sum of each anti-magical sequence is equal to the sum of all the rows, all the columns, and the two diagonals:

Sequence sum = Rows + Columns + Diagonals . \text{Sequence sum} = \text{Rows} + \text{Columns} + \text{Diagonals}.

Let D D be the sum of the diagonal sums. We have two cases to consider, one for each anti-magical sequence:

k 4 + k 3 + k 2 − 1 = 2 ( k 4 + k 2 2 ) + D k 4 + k 3 + k 2 − 1 = k 4 + k 2 + D D = k 3 − 1 , \begin{aligned} k^4 + k^3 + k^2 - 1 &= 2 \left ( \dfrac{k^4 + k^2}{2} \right ) + D \\ k^4 + k^3 + k^2 - 1 &= k^4 + k^2 + D \\ D &= k^3 - 1, \\ \end{aligned}

and

k 4 + k 3 + k 2 + 2 k + 1 = 2 ( k 4 + k 2 2 ) + D k 4 + k 3 + k 2 + 2 k + 1 = k 4 + k 2 + D D = k 3 + 2 k + 1. \begin{aligned} k^4 + k^3 + k^2 + 2k + 1 &= 2 \left ( \dfrac{k^4 + k^2}{2} \right ) + D \\ k^4 + k^3 + k^2 + 2k + 1 &= k^4 + k^2 + D \\ D &= k^3 + 2k + 1. \end{aligned}

Therefore, there are n = 2 n = 2 possible values for the sum of the diagonal sums in a k × k k \times k anti-magical square, which are d 1 = k 3 − 1 d_1 = k^3 - 1 and d 2 = k 3 + 2 k + 1. d_2 = k^3 + 2k + 1. We have

n ( d 1 + d 2 ) = 2 ( ( k 3 − 1 ) + ( k 3 + 2 k + 1 ) ) = 2 ( 2 k 3 + 2 k ) = 4 k ( k 2 + 1 ) . n(d_1 + d_2) = 2((k^3 - 1) + (k^3 + 2k + 1)) = 2(2k^3 + 2k) = 4k(k^2 + 1).

Substituting in k = 15 k = 15 yields 4 ( 15 ) ( 1 5 2 + 1 ) = 13560 . 4(15)(15^2 + 1) = \boxed{13560}.

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