===Open Problem #2===

Introduction: For the sake of ease, let's label the squares positions a to i from the top left to the bottom right. Considering there are two diagonals, three verticals, and three horizontals, there are 8 numbers which must be consecutive in order for this to happen. Say n is the smallest of these numbers. The sum of these numbers is 8n+28. We know that the sum must be more than 90, since that is the sum without the diagonals.

Proving the sum of the sums must be less than 135: The sum of the diagonals is a+c+2e+g+i. To prove that the sum must be less than 135, we need to prove that that is less than 45 which is the sum of all 9 letters a through i. Cancelling numbers that appear on both sides we must prove that e must be less than b+d+f+h. Conveniently, none of the nine letters are the same, and are all positive integers less than or equal to 9. Thus 9 is the largest possible value for e. The smallest sum of four letters possible is 10. Thus e < b+d+f+h. So the sum of all eight sums must be less than 135.

Inequalities: 90 < T < 135

90 < 8n+28 < 135

62 < 8n < 107

Since n is an integer, we know that 64 < 8n < 104

8 < n < 13

Conclusion: Based on the above calculations, we only need to worry about chains of 8 numbers with the starting number between 8 and 13 inclusively or

8, 9, 10, 11, 12, 13, 14, 15

9, 10, 11, 12, 13, 14, 15, 16

10, 11, 12, 13, 14, 15, 16, 17

11, 12, 13, 14, 15, 16, 17, 18

12, 13, 14, 15, 16, 17, 18, 19

13, 14, 15, 16, 17, 18, 19, 20

are the only possibilities.

I've been thinking about Mike's solution for this whole afternoon, and it seems like I've found another way, which can be generalized for all $n$, to reduce "the chain of numbers that we need to worry about" to 2.

(Details of detailed proof when $k=3$ is in Mike Harding's comment, and is highly recommended for people who don't want to read the generalization.)

According to Mike's solution, in a $k^2$ square, there must be $2k+2$ consecutive values for the $k^2$ square to be anti-magic. Let $n$ be the smallest value, then the $2k+2$ consecutive values are $n,n+1,...,n+2k+1$, and the sum of those consecutive values are $2kn+2n+(2k+1)(k+1)$. The sum of the $k^2$ numbers inside is $\frac {k^2(k^2+1) }{2 }$. Also, for a chain to be proved as incorrect, either $(2k+1)n-k^2(k^2+1)+(2k+1)(k+1)<2n+1$ or $(2k+1)n-k^2(k^2+1)+(2k+1)(k+1)>2n+4k+2$ must be true.

Case 1: $(2k+1)n-k^2(k^2+1)+(2k+1)(k+1)<2n+1$

This can be reduced to $0>k(2n-k^3+k+3)$ or $2n<k^3-k-3$.

Case 2: $(2k+1)n-k^2(k^2+1)+(2k+1)(k+1)>2n+4k+2$

This can be reduced to $k(2n-k^3+k-1)>1$ or $2n>k^3-k+1+\frac { 1 }{ k }$.

So the values that can be considered "eligible" (I don't even know what word to use right now, sorry) are $k^3-k-3\le2n\le k^3-k+1+\frac { 1 }{ k }$.

Note that $k^3-k$ is divisible by 2 and $\frac{1}{k}$ is not an integer, so we can say that $\frac{k^3-k-2}{2} \le n \le \frac{k^3-k}{2}$.

And as a result, the two chains of $2k+2$ numbers that we are looking for are $\frac { k^3-k-2 }{ 2 } ,\frac { k^3-k }{ 2 } ,...,\frac { k^3+3k }{ 2 }$ and $\frac { k^3-k }{ 2 } ,\frac { k^3-k+2 }{ 2 } ,...,\frac { k^3+3k+2 }{ 2 }$.

For $k=3$, the two chains that we are looking for are $11, 12, 13, 14, 15, 16, 17, 18$ and $12, 13, 14, 15, 16, 17, 18,19$.

I started looking for a reason no 3x3 antimagic square exists. To be more specific, I started looking at the relations between even and odd numbers. We need to place the numbers 1 through 9. Of those numbers, 4 are even and 5 are odd. Furthermore, there are a total of 8 sums: 3 rows, 3 columns, and 2 diagonals. These sums must be consecutive numbers, so 4 of these sums are even and 4 are odd.

I started looking for all the ways of placing 4 even and 5 odd numbers in a 3x3 square and calculated how many of the 8 sums would evaluate to some even number. If exactly 4 of these sums were even, it's a valid way of placing even and odd numbers. I found that only 17 ways of ordering even and odd numbers were valid. After taking out rotations, only 5 ways were left. There are the 5 ways:

$\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$

0 represents some even number and 1 represents some odd number. Source code I used for finding those orderings

I also did some math on how many ways there are to arrange the numbers 1 through 9. There are 9! = 362,880 ways. Furthermore, there are $\binom{9}{4} = 126$ ways of placing 4 even numbers and 5 odd numbers, and 109 of those don't produce the required 4 even and 4 odd sums. So $362,880 / 126 * 109 = 313,920$ of the orderings are impossible.

($LaTeX$ is up, please let me know if there's still any errors.)

As usual, we label the squares as such:

$\begin{array}{c|c|c} a & b & c \\ \hline d & e & f \\ \hline g & h & i \end{array}$

Mike and Steven Jim have established that only two sequences are possible for a 3x3 anti-magic square: $11, 12, \dots, 18$ and $12, 13, \dots, 19.$ Let’s focus on the first possible "anti-magical" sequence. The sum of all the numbers in the sequence is equal to 116. This is equal to the combined sum of the numbers in the three rows, the numbers in the three columns, and the numbers in the two diagonals. Since both the union of all the rows and the union of all the columns is $\{1, 2, \dots, 9\},$ their sum is equal to 45, and the diagonal sum is

$(a + e + i) + (c + e + g) = 116 - 2(45) = 26.$

Now, consider the sum of the numbers from the middle column and middle row (the "middle sums"). It’s equal to $(b + e + h) + (d + e + f).$ Summing this with the diagonal sum gives us

$(a + e + i) + (c + e + g) + (b + e + h) + (d + e + f) = (a + b + c + d + e + f + g + h + i) + 3e = 3e + 45.$

Plugging in our value for the diagonal sum gives us $(b + e + h) + (d + e + f) = 3e + 45 - 26 = 3e + 19.$

There are only two pairs of distinct integers in the range $[11, 18]$ that sum to 26: $(11, 15)$ and $(12, 14).$ This means that one diagonal sums to one of the numbers in a pair, and the other diagonal sums to the other. We can assign a value to the middle sums by noting that it must be one more than a multiple of 3, since it's equal to $3e + 19 \equiv 1 \! \pmod{3}.$ Here we must do some casework:

• If we chose $(11, 15)$ for the diagonals, we are left with 12, 13, 14, 16, 17, 18 for the middle sums. We must choose two integers that sum to a number that is two more than a multiple of three. This means the possible middle sums are $(12, 13), (12, 16), (13, 18), (14, 17), (16, 18).$

• If we chose $(12, 14)$ for the diagonals, we are left with 11, 13, 15, 16, 17, 18 for the middle sums. Using the same criteria as the previous case, the possible middle sums are $(11, 17), (13, 15), (13, 18), (15, 16), (16, 18).$

For each possible middle sum pair, we can find the value of $e$ i.e. the middle number by using middle sum = $3e + 19.$ For example, if we chose $(11, 15)$ for the diagonals and $(14, 17)$ for the middles, then $e = \frac{14 + 17 - 19}{3} = 4.$ If we do this for every possible middle sum pair, we find that if there exists an anti-magic square whose sequence is 11 to 18, then the middle number $e$ is between 2 and 5 inclusive.

Applying the same procedure to the sequence $12, 13, \dots, 19,$ we can conclude that $e \in [5, 8]$ in this case. The middle number cannot be 1 or 9.

@Steven Jim here it is. (Sorry it took so long, I hurt my hand recently so typing is much harder. Also, this post is a continuation of my first one.)

The possible values for pairs of diagonals and pairs of middles, as well as their corresponding values for $e,$ are shown here:

$\begin{array}{c|c|c} \text{Diagonals} & \text{Middles} & e \\ \hline (11, 15) & (12, 13) & 2 \\ & (12, 16) & 3 \\ & (13, 18) & 4 \\ & (14, 17) & 4 \\ & (16, 18) & 5 \\ \hline (12, 14) & (11, 17) & 3 \\ & (13, 15) & 3 \\ & (13, 18) & 4 \\ & (15, 16) & 4 \\ & (16, 18) & 5 \\ \hline (15, 19) & (12, 14) & 5 \\ & (12, 17) & 6 \\ & (13, 16) & 6 \\ & (14, 18) & 7 \\ & (17, 18) & 8 \\ \hline (16, 18) & (12, 14) & 5 \\ & (12, 17) & 6 \\ & (13, 19) & 7 \\ & (14, 15) & 6 \\ & (15, 17) & 7 \end{array}$

For each case, we can calculate $a + c + g + i,$ $b + h,$ and $d + f$ by subtracting $2e$ from the diagonal sum and $e$ from each individual middle sum. WLOG we can assume that the first number in each pair of middle sums equals $b + e + h$ and the other equals $d + e + f.$ To simplify the proceeding explanation, let's focus on one pair of diagonal and middle sums - $(11, 15)$ and $(13, 18).$ The corner number sum is

$a + c + g + i = 26 - 2e = 26 - 2(4) = 18,$

and we also have

$b + h = 13 - e = 13 - 4 = 9 \\ d + f = 18 - e = 18 - 4 = 14.$

Now we can use these values to calculate the sum of the "outer rows," $(a + b + c) + (g + h + i),$ and the sum of the "outer columns," $(a + d + g) + (c + f + i).$

$\begin{aligned} (a + b + c) + (g + h + i) &= (a + c + g + i) + (b + h) = 18 + 9 = 27 \\ (a + d + g) + (c + f + i) &= (a + c + g + i) + (d + f) = 18 + 14 = 32. \end{aligned}$

The remaining sum values we haven't used yet for this case are 12, 14, 16, and 17. We must assign these values to the outer rows and columns such that the above relations are true. However, it is fairly easy to see that no assignment of these four integers to $(a + b + c, g + h + i, a + d + g, c + f + i)$ will satisfy the above requirements. Thus, if an anti-magical square of order 3 exists, then the diagonal and middle sums cannot be $(11, 15)$ and $(13, 18),$ respectively.

We can use the above method on the other 19 possible cases, checking to see if we can assign the remaining sum values to the outer rows and columns. These eight cases do not work out:

$\begin{array}{c|c} \text{Diagonals} & \text{Middles}\\ \hline (11, 15) & (12, 16) \\ & (13, 18) \\ \hline (12, 14) & (13, 15) \\ & (16, 18) \\ \hline (15, 19) & (12, 17)\\ & (14, 18) \\ \hline (16, 18) & (12, 14) \\ & (15, 17) \end{array}$

The other twelve have valid sum value assignments, so we would have to directly check them to verify that they don't work.

EDIT: After reviewing the even-odd placements that Stefan provided us in a different comment, we were able to remove all of the $(12, 14)$ and $(16, 18)$ diagonal sum cases, leaving us with only 6 cases left to check:

$\begin{array}{c|c} \text{Diagonals} & \text{Middles}\\ \hline (11, 15) & (12, 13) \\ & (14, 17) \\ & (16, 18) \\ \hline (15, 19) & (12, 14)\\ & (13, 16) \\ & (17, 18) \end{array}$

@Steven Jim You asked if I was able to proof it's impossible to have a 3x3 antimagic square where at least one diagonal is even. I think I've managed to do it.

Like in previous work, we use the following naming scheme:

$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$

From previous work we know the following:

• We need to place the number 1, 2, 3, ..., 9. Five of these numbers are odd, and four are even
• The combined sum of all the rows, all the columns, and the two diagonals must add up to either 116 or 124. These are both even numbers
• Of the eight sums for the rows, columns, and diagonals, 4 must be even and 4 odd

Since the combined sum of all rows, columns, and diagonals must be even, we can calculate the following: $(a + b + c) + (d + e + f) + (g + h + i) + (a + d +g) + (b + e + h) + (c + f + i) + (a + e + i) + (c + e + g) = 3a + 2b + 3c + 2d + 4e + 2f + 3g + 2h + 3i \equiv a + c + g + i \equiv 0\ (\textrm{mod}\ 2)$

So $a + c + g + i$, the sum of all the corners, must be equal to some even number. The only way that's true is if exactly 0, exactly 2, or exactly 4 of the corners are odd. Let's look at all these cases, and check if it's possible to have at least one diagonal with an even sum.

Case 1: none of the corners are odd

In this case, the four even numbers must all go into the corners, and the five odd numbers go in the remaining spots. This leads to the following 3x3 matrix. 1 means 'some odd number' and 0 means 'some even number': $\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ The sums of both diagonals are odd, so in this case it's impossible to have at least one even diagonal.

Case 2: all 4 corners are odd

If all corners are equal to an odd number, and at least one diagonal must be equal to some even number, the center must be some even number. Due to rotational symmetry, it doesn't matter where the last odd number gets placed. This leads to the following 3x3 matrix: $\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$

This matrix has 2 even rows, 2 even columns, and 2 even diagonals. The remaining 2 sums are odd. This isn't equal to the required 4 even and 4 odd sums. Therefore case 2 also can't help us create an antimagic square with at least one diagonal with an even sum.

Case 3: exactly 2 corners are odd

There are two ways of placing 2 odd numbers in the corners. There are two ways of placing these odd numbers:

Case 3 sub-case 1: the odd numbers are opposite to each other

To make sure there's at least one even diagonal, the center must have some even number. The 3x3 matrix now looks like this: $\begin{bmatrix} 0 & ? & 1 \\ ? & 0 & ? \\ 1 & ? & 0 \end{bmatrix}$ Due to rotational and reflection symmetry, it doesn't matter where we place the fourth even number. So we might fill out the matrix entirely like this: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ This matrix has 2 even rows, 2 even columns, and 2 even diagonals. The remaining 2 sums are odd. This isn't equal to the required 4 even and 4 odd sums. Therefore case 3 sub-case 1 also can't help us create an antimagic square with at least one diagonal with an even sum.

Case 3 sub-case 2: the odd numbers are adjacent to each other

To make sure there's at least one even diagonal, the center must have some odd number number. The 3x3 matrix now looks like this: $\begin{bmatrix} 1 & ? & 0 \\ ? & 1 & ? \\ 1 & ? & 0 \end{bmatrix}$ There are 4 ways of placing the remaining 2 even and 2 odd numbers, after taking out reflections: $\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ The first matrix only has 2 even diagonals. The other 3 matrix has 2 even rows, 2 even columns, and 2 even diagonals. None of these are the required 4 even and 4 odd sums. Therefore case 3 subcase 2 also can't help us create an antimagic square with at least one diagonal with an even sum.

Since all of the cases are dead ends, it's impossible to have a 3x3 antimagic square with at least one even diagonal. $\square$

I've been thinking about the last 6 cases Steven Yuan presented.

$\begin{array}{c|c|c} \text{Diagonals} & \text{Middles} & e \\ \hline (11,15) & (12,13) & 2 \\ & (14,17) & 4 \\ & (16,18) & 5 \\ \hline (15,19) & (12,14) & 5 \\ & (13,16) & 6 \\ & (17,18) & 8 \end{array}$

So far the best approach to prove all of these cases don't result in 3x3 antimagic squares was to look at many different cases, but I think I've found a shorter proof.

Like usual, I use the following naming convention:

$\begin{array}{c|c|c} a & b & c \\ \hline d & e & f \\ \hline g & h & i \\ \end{array}$

Case 1: diagonals (11, 15) and middle sums (12, 13)

In this case, e is 2. If we subtract e from all the diagonals and middles, and look at all the ways we can add up two of the remaining 8 digits to get those remainders, we get the following options:

$\begin{array}{c|c|c|c} \text{Diagonal 1: } 9 & \text{Diagonal 2: } 13 & \text{Middle 1: } 10 & \text{Middle 2:} 11 \\ \hline (1,8) & (4, 9) & (1, 9) & (3, 8) \\ (3, 6) & (5,8) & (3, 7) & (4, 7) \\ (4, 5) & (6,7) & (4, 6) & (5, 6) \end{array}$

We must select one option from every column, and we need to select every digit exactly once. There are three ways to do that:

$\begin{array}{c|c|c|c} \text{Diagonal 1: } 9 & \text{Diagonal 2: } 13 & \text{Middle 1: } 10 & \text{Middle 2:} 11 \\ \hline (1,8) & (4, 9) & (3, 7) & (5, 6) \\ (3, 6) & (5,8) & (1, 9) & (4, 7) \\ (4, 5) & (6, 7) & (1, 9) & (3, 8) \end{array}$

To create an antimagic square, the outer sums, $(a + b + c)$, $(g + h + i)$, $(a + d + g)$, and $(c + f + i)$ must add up to the remaining four sums: 14, 16, 17, and 18. For every sum, we need to pick exactly 1 digit from every diagonal, and exactly 1 digit from one of the middles.

Now try to make 14 by picking exactly one digit from each diagonal and one from a middle sum for the first option. It becomes quickly clear that it's not possible to do that. It's also not possible to make 14 with the second option. It is possible to do in the third option $(4 + 7 + 3 = 14)$, but it's impossible to create a 16 with the third option. Therefore the first case, $(11, 15)$ for the diagonals and $(12, 13)$ for the middle sums won't work.

Case 2: diagonals (11, 15) and middle sums (14, 17)

In this case, e is 4. Again, the ways to add up the remaining digits to get the sums:

$\begin{array}{c|c|c|c} \text{Diagonal 1: } 7 & \text{Diagonal 2: } 11 & \text{Middle 1: } 10 & \text{Middle 2:} 13 \\ \hline (1,6) & (2, 9) & (1, 9) & (5, 8) \\ (2, 5) & (3,8) & (2, 8) & (6, 7) \\ & (5,6) & (3, 7) & \end{array}$

The two ways to select one from each column, together with the reason why that case won't work:

$\begin{array}{c|c|c|c|c} \text{Diagonal 1: } 9 & \text{Diagonal 2: } 13 & \text{Middle 1: } 10 & \text{Middle 2:} 11 & \text{Reason it won't work} \\ \hline (2,5) & (3, 8) & (1, 9) & (6, 7) & \text{Impossible to make 13} \\ (1, 6) & (2,9) & (3, 7) & (5, 8) & \text{Impossible to make 12} \end{array}$

Case 3: diagonals (11, 15) and middle sums (16, 18)

In this case, e is 5. In this case, there's only one way of selecting the digits:

$\begin{array}{c|c|c|c} \text{Diagonal 1: } 6 & \text{Diagonal 2: } 10 & \text{Middle 1: } 11 & \text{Middle 2:} 13 \\ \hline (2,4) & (1, 9) & (3, 8) & (6, 7) \end{array}$

It's also possible to create all of the remaining sums: 12, 13, 14, and 17. For all of these numbers there's only one way to make them, and when we look closer at the 12, and 13, it becomes clear why this case won't create an antimagic square.

$12 = 4 + 1 + 7 \\ 13 = 4 + 1 + 8$

The sums for 12 and 13 are both trying to grab the 1 and 4 from the diagonals, but only one number is allowed to do that.

Case 4: diagonals (15, 19) and middle sums (12, 14)

In this case, e is 5. In this case, there's only one way of selecting the digits:

$\begin{array}{c|c|c|c} \text{Diagonal 1: } 10 & \text{Diagonal 2: } 14 & \text{Middle 1: } 7 & \text{Middle 2:} 9 \\ \hline (1,9) & (6, 8) & (3, 4) & (2, 7) \end{array}$

It's also possible to create all of the remaining sums: 13, 16, 17, and 18. For all of these numbers there's only one way to make them, and when we look closer at the 13, and 16, it becomes clear why this case won't create an antimagic square.

$13 = 1 + 8 + 4 \\ 16 = 1 + 8 + 7$

The sums for 13 and 16 are both trying to grab the 1 and 8 from the diagonals, but only one number is allowed to do that.

Case 5: diagonals (15, 19) and middle sums (13, 16)

In this case, e is 6. Similar to what was done in case 2:

$\begin{array}{c|c|c|c|c} \text{Diagonal 1: } 9 & \text{Diagonal 2: } 13 & \text{Middle 1: } 7 & \text{Middle 2:} 10 & \text{Reason it won't work} \\ \hline (2,7) & (5, 8) & (3, 4) & (1, 9) & \text{Impossible to make 12} \\ (1, 8) & (4, 9) & (2, 5) & (3, 7) & \text{Impossible to make 18} \end{array}$

Case 6: diagonals (15, 19) and middle sums (17, 18)

In this case, e is 8. Similar to what was done in case 5:

$\begin{array}{c|c|c|c|c} \text{Diagonal 1: } 7 & \text{Diagonal 2: } 11 & \text{Middle 1: } 9 & \text{Middle 2:} 10 & \text{Reason it won't work} \\ \hline (1,6) & (2, 9) & (4, 5) & (3, 7) & \text{Impossible to make 16} \\ (2, 5) & (4, 7) & (3, 6) & (1, 9) & \text{Impossible to make 14} \\ (3, 4) & (5, 6) & (2, 7) & (1, 9) & \text{Impossible to make 13} \end{array}$

All 6 cases are impossible, therefore there is no 3 by 3 antimagic square. $\square$

I did some googling, and found a few links to work done by other people, so we already have a start:

• The Anti-Magic Square Project
• Anti-Magic Squares of Even Orders

We know that the AMS must have the integers from 1-9. Let us formalize this as follows. A1,A2,A3 B1,B2,B3 C1,C2,C3 A2,B1,C2,B3 are edges A1,A3,C1,C3 are corners, B2 is the center.

A1+B1+C1=R1 A2+B2+C2=R2 A3+B3+C3=R3 A1+A2+A3=A B1+B2+B3=B C1+C2+C3=C A1+B2+C3=X C1+B2+A3=Y R1+R2+R3=A+B+C=45 A1 through C3 are distinct integers from 1-9. R1+R2+R3+A+B+C+X+Y=(N+1)N/2-(M-1)M/2, where M is the lowest sum and N is the highest sum. Notice that N=M+7. 45+45+A1+C1+A3+C3+2B2=(M+8)(M+7)/2-(M-1)M/2=(M^2+15M+56-M^2+M)/2=(16M+56)/2=8M+28 62+A1+C1+A3+C3+2B2=8M This tells us that A1+C1+A3+C3 is even, meaning the corners must have 0, 2, or 4 odd numbers. If A1=2,C1=3,A3=4,C3=5,B2=1, then the lowest possible value of M is ceil((62+2+3+4+5+2)/8)=10. If B2=9,A1=8,A3=7,C1=6,C3=5, then the highest possible value of M is floor((62+18+8+7+6+5)/8)=13.

This means the lowest of the sums must be either 10,11,12, or 13, and the highest of the sums must be either 17,18,19,20.

Let's look at the center. We see that it creates four pairs of numbers whose sums (R2,B,X,Y) must be distinct, such that 20>=S_n>= 10. This means the minimum sum of the pair sums is 10+11+12+13=50. This means that A1+B1+C1+C2+C3+B3+A3+A2+4B2>=50, or 45+3B2>=50, or 3B2>=5, or B2>=2, meaning B2 cannot be 1.

Okay. I've just had an idea on proving if $n^2$ anti-magic squares do exist for $n \ge 4$. So for all even $n$, Stefan Van der Waal found out a link which shows us the proof. For odd $n$, if we can show that a $7^2$ anti-magic square does exist, we may prove that $3$ is the maximum value so that no anti-magic square exists. However, if we can prove that a $7^2$ anti-magic square does not exist, we can prove that the general question is only true when $n=4k+1$, with $k\ge1$. What are your thoughts?

I just had a thought, don't know how useful it might be but it could make an interesting problem here on Brilliant. This concerns the possible diagonal sums in a $k \times k$ anti-magic square, where $k$ is an odd number greater than 3, and it is an extension of the work done for the 3x3 case.

We've established that for any $k,$ there are only two possible sum sequences for the $k \times k$ anti-magic square: $\frac{k^3 - k - 2}{2}, \frac{k^3 - k}{2}, \dots, \frac{k^3 + 3k}{2}$ and $\frac{k^3 - k}{2}, \frac{k^3 - k + 2}{2}, \dots, \frac{k^3 + 3k + 2}{2}.$ Let's focus on the first sequence. The sum of the numbers in this sequence is

$\dfrac{1}{2}(2k + 2) \left ( \dfrac{k^3 - k - 2}{2} + \dfrac{k^3 + 3k}{2} \right ) = (k + 1)(k^3 + k - 1) = k^4 + k^3 + k^2 - 1.$

The sum of all the integers from 1 to $k^2$ inclusive is simply $\frac{k^4 + k^2}{2}.$ This is equal to both the sum of all the rows equals the sum of all the columns in the square, since both cases cover all the integers on the grid. Let $d_1, d_2$ be the diagonal sums, with $d_1 < d_2.$ We can write

$\begin{aligned} \text{Sum of sequence} &= \text{Rows} + \text{Columns} + \text{Diagonals} \\ k^4 + k^3 + k^2 - 1 &= 2 \left ( \dfrac{k^4 + k^2}{2} \right ) + d_1 + d_2 \\ k^4 + k^3 + k^2 - 1 &= k^4 + k^2 + d_1 + d_2 \\ d_1 + d_2 &= k^3 - 1. \\ \end{aligned}$

Since $k$ is odd, $k^3 - 1$ is even, and $\frac{k^3 - 1}{2}$ is an integer. Clearly, $\frac{k^3 - 1}{2} < \frac{k^3 + 3k}{2},$ and we also have $\frac{k^3 - 1}{2} > \frac{k^3 - k - 2}{2},$ as that reduces to $k > -1.$ Thus, $\frac{k^3 - 1}{2}$ is part of the sequence of sums.

Finally, the number of possible values of $d_1,$ which is also equal to the number of pairs $(d_1, d_2),$ is just $\frac{k^3 - 1}{2} - \frac{k^3 - k - 2}{2} = \frac{k + 1}{2},$ since we must exclude $\frac{k^3 - 1}{2}$ as it will cause $d_1 = d_2,$ and any other value less than that is permissible. Therefore, there are $\frac{k + 1}{2}$ possible diagonal sum pairs for the first sequence of sums.

Applying the same logic to the second sequence, which starts at $\frac{k^3 - k}{2},$ we find another $\frac{k + 1}{2}$ diagonal sum pairs. The total number of possible diagonal sum pairs is $k + 1.$

Okay... @Steven Yuan @Stefan van der Waal @Mike Harding I've been trying to do some brute force with the remaining 6 cases, but I just found out that there are too many sub-cases to consider.

First of all, yes, I can find the values from $a$ to $i$. The problem is: For every pair of diagonal and middle sums, there seem to be at least 4 sub-cases. Also, there are 8 ways to rearrange (without rotation, hell) the 3x3 board so that the diagonal and middle sums remain the same, yet the other sums. So... Yes... We will have at least 192 sub-cases to consider. That's not how I want to finish the proof, tbh. So... We do need a way to throw more cases away. Ideas?

There are 3120 unique 3x3 heterosquares.