A special semimagic square

[ArkadWesolowski]

What is so special about the following semimagic square?
105, 406, 19110;
19306, 300, 15;
210, 18915, 496;
This semimagic square is composed of nine distinct triangular numbers.
The magic sum S is 19621.
One of the diagonals sums to 19620 = S - 1.
Is it possible to make a 3 X 3 semimagic square whose rows, columns, and just one of the two main diagonals sum to the same number and whose entries are all distinct triangular numbers?

[science_man_88]

Originally Posted by ArkWesolowski

What is so special about the following semimagic square?
105, 406, 19110;
19306, 300, 15;
210, 18915, 496;
This semimagic square is composed of nine distinct triangular numbers.
The magic sum S is 19621.
One of the diagonals sums to 19620 = S - 1.
Is it possible to make a 3 X 3 semimagic square whose rows, columns, and just one of the two main diagonals sum to the same number and whose entries are all distinct triangular numbers?

okay so let's break this down you want:
1) all triangular numbers
2) all but one diagonal summing to the same number
3) the result to be a 3 by 3 square

1.1) triangular numbers are of form

2.1) as there are 6 connections to be made across the square you need a number with at least 5 partitions that share a number with at least one other.

4) so what would a sum that worked look like ( that way if we code it we might be able to code out sums that don't work). The answer is where means sum. Suppose we want to have a lower search limit ( a number we can't go below and find a solution). what could we figure out about it relatively naively well for one we can say it can't be the sum of the first three as those sum to the fourth so there's only one possible representation as the sum of three so that's out. we can then go on to say well how can we every get more 4 representations? Well we can have more than 4 representations when the number of ways of choosing k from a group of n ( the binomial coefficient expressed as ) is greater than 4. and we can work it out even further when there can be the correct number of overlaps etc. anyways I'm blabbing on without knowing what terms I can use and make sense. edit: I fail the reading test as your wording suggest you posted the answer before asking the question.

[science_man_88]

I guess one semi interesting thing for me is that all the triangular numbers ending with 6 if reduced by 1 make every element divisible by 15 but I'm easily amazed.

[wblipp]

I think that you can derive this from a semi-magic square of odd squares. A quick online search looks like the magic squares of squares puzzle is still unsolved and I didn't find anything about semi-magic squares of squares.

[ArkadWesolowski]

For more details, see Problem 63.

[Anonuser]

Originally Posted by ArkWesolowski

What is so special about the following semimagic square?
105, 406, 19110;
19306, 300, 15;
210, 18915, 496;
This semimagic square is composed of nine distinct triangular numbers.
The magic sum S is 19621.
One of the diagonals sums to 19620 = S - 1.
Is it possible to make a 3 X 3 semimagic square whose rows, columns, and just one of the two main diagonals sum to the same number and whose entries are all distinct triangular numbers?

I think the answer is yes. A few examples:

25894806 18547095 4846941
589155 13825911 34873776
22804881 16915836 9568125

45453345 13356696 3810180
233586 10371735 52014900
16933290 38891790 6795141

92759010 18547095 4846941
589155 13825911 101737980
22804881 83780040 9568125

92759010 18547095 16915836
589155 25894806 101737980
34873776 83780040 9568125

92759010 22804881 16915836
4846941 25894806 101737980
34873776 83780040 13825911

409080106 120210265 34291621
2102275 93345616 468134101
152399611 350026111 61156270

834831091 205243930 152242525
43622470 233053255 915641821
313863985 754020361 124433200

1730278551 98343300 52474890
17638830 92827125 1770630786
133179360 1689926316 57991065