Has this approach to the 3x3 Magic Square of Squares problem already been tried?

[Simpson17866]

I found a formula for a magic square that guarantees 5 out of the 9 numbers to be perfect square numbers (the 4 corners and the center) for any value of (x,y)

 

 

And before Microsoft Excel succumbed to rounding errors, I found four specific values of (x,y) which bring this up to 6 out of 9 (the 4 corners, the center, and one of the 4 sides)

 

(1,3)

with a central value of 125^2 = 15,625

and a right-hand value of 95^2 = 9,025

 

(1,10)

with a central value of 1,105^2 = 1,221,025

and a right-hand value of 529^2 = 279,841

 

(1,59)

with a central value of 35,405^2 = 1,253,514,025

and an upper value of 2,831^2 = 8,014,561

 

(3,41)

with a central value of 86,125^2 = 7,417,515,625

and a bottom value of 108,455^2 = 11,762,487,025

 

Have other mathematicians already tried magic squares of this form? If so, has this formula already been ruled out as being incapable of generating 9 perfect square numbers out of 9?

Edited July 1, 2017 by Simpson17866

[Raider5678]

I found a formula for a magic square that guarantees 5 out of the 9 numbers to be perfect square numbers (the 4 corners and the center) for any value of (x,y)

 

Screen Shot 2017-07-01 at 1.36.25 PM.png

 

And before Microsoft Excel succumbed to rounding errors, I found four specific values of (x,y) which bring this up to 6 out of 9 (the 4 corners, the center, and one of the 4 sides)

 

(1,3)

with a central value of 125^2 = 15,625

and a right-hand value of 95^2 = 9,025

 

(1,10)

with a central value of 1,105^2 = 1,221,025

and a right-hand value of 529^2 = 279,841

 

(1,59)

with a central value of 35,405^2 = 1,253,514,025

and an upper value of 2,831^2 = 8,014,561

 

(3,41)

with a central value of 86,125^2 = 7,417,515,625

and a bottom value of 108,455^2 = 11,762,487,025

 

Have other mathematicians already tried magic squares of this form? If so, has this formula already been ruled out as being incapable of generating 9 perfect square numbers out of 9?

 

 

I don't really get it.

 

Are you basically choosing a number for x, and then no matter what x is it will be a square inside of at least 6 of them?

 

What if you just put x^2 in all of them?

[Simpson17866]

I don't really get it.

 

Are you basically choosing a number for x, and then no matter what x is it will be a square inside of at least 6 of them?

That's the plan.

 

The basic form of any magic square is

 

E+n; E-n-m; E+m

 

E-n+m; E; E+n-m

 

E-m; E+n+m, E-n

 

For any arbitrary values of (E, n, m), the 3 rows, the 3 columns, and the 2 diagonals will all add up to the same "magic sum" of 3E.

 

Setting E=5, n=1, and m=3 gives the "classic" magic square that uses each digit from 1 to 9 exactly once each:

 

6; 1; 8

7; 5; 3

2; 9; 4 (magic sum: 3*5 = 15)

 

This contains 3 perfect square numbers: 1; 4; and 9.

 

Whereas, if you set E=25, n=11, and m=24, then you get a magic square that contains 4 square numbers instead of just 3:

 

36; -10; 49

38; 25; 12

1; 60; 14

 

The challenge is to find a magic square that uses 9 different square numbers instead of just 3 or 4. So far, the best anybody's managed is

 

8/8 sums the same, 7/9 square numbers

9/9 square numbers, 7/8 sums the same

 

Nobody knows whether or not it's possible to make it all the way to 8/8 sums being the same by using 9/9 square numbers.

 

If it is possible, though, then the numbers would have to be horrendously large (the minimum for any one of them would have to be 10,000,000^2, and it's possible that googol^2 might not be big enough)

 

What if you just put x^2 in all of them?

That is considered cheating

[Raider5678]

That's the plan.

 

The basic form of any magic square is

 

E+n; E-n-m; E+m

 

E-n+m; E; E+n-m

 

E-m; E+n+m, E-n

 

For any arbitrary values of (E, n, m), the 3 rows, the 3 columns, and the 2 diagonals will all add up to the same "magic sum" of 3E.

 

Setting E=5, n=1, and m=3 gives the "classic" magic square that uses each digit from 1 to 9 exactly once each:

 

6; 1; 8

7; 5; 3

2; 9; 4 (magic sum: 3*5 = 15)

 

This contains 3 perfect square numbers: 1; 4; and 9.

 

Whereas, if you set E=25, n=11, and m=24, then you get a magic square that contains 4 square numbers instead of just 3:

 

36; -10; 49

38; 25; 12

1; 60; 14

 

The challenge is to find a magic square that uses 9 different square numbers instead of just 3 or 4. So far, the best anybody's managed is

 

8/8 sums the same, 7/9 square numbers

9/9 square numbers, 7/8 sums the same

 

Nobody knows whether or not it's possible to make it all the way to 8/8 sums being the same by using 9/9 square numbers.

 

If it is possible, though, then the numbers would have to be horrendously large (the minimum for any one of them would have to be 10,000,000^2, and it's possible that googol^2 might not be big enough)

 

That is considered cheating

So you do it by having all three E,N,and M values inside the equation?

I'm confused.

 

So the challenge is, have 9 boxes, with different formulas in each.

The formulas use X and Y.

For any value of X and Y, you have to make it equal a square in each box.

And then on top of that, have all the sums equal for all horizontal paths, all diagonal paths, and all vertical paths?

[Simpson17866]

So you do it by having all three E,N,and M values inside the equation?

I'm confused.

 

So the challenge is, have 9 boxes, with different formulas in each.

The formulas use X and Y.

For any value of X and Y, you have to make it equal a square in each box.

And then on top of that, have all the sums equal for all horizontal paths, all diagonal paths, and all vertical paths?

You don't have to write it in terms of X and Y, that was just how my personal approach developed.

 

The general form is just

 

E+n = a^2

E-n-m = b^2

E+m = c^2

 

E-n+m = d^2

E = e^2

E+n-m = f^2

 

E-m = g^2

E+n+m = h^2

E-n = i^2

 

And I happened to find one specific set of partial solutions defined by two variables that I happened to designate as x and y. Other partial solutions wouldn't take this specific form, and a lot of them are probably better than mine.

[Simpson17866]

So I've known ever since I discovered this formula that it was an extremely special case that skipped the vast majority of the magic squares that have square numbers in the center and the corners, but I'd thought that a formula that would look at all of the magic squares with square numbers there would be too complicated to use.

 

I just found a formula that looks almost exactly the same, but that I believe covers every single magic square that could possibly have 5 square numbers in the corners and center

 

E+n = (2p^2 + 4pq + q^2)^2 (2r^2 + 2rs + s^2)^2

E-n = (2p^2 - q^2)^2 (2r^2 + 2rs + s^2)

 

E = (2p^2 + 2pq + q^2)^2 (2r^2 + 2rs + s^2)^2

 

E+m = (2p^2 + 2pq + q^2)^2 (2r^2 + 4rs + s^2)^2

E-m = (2p^2 + 2pq + q^2)^2 (2r^2 - s^2)^2

 

For which my original formula was the special case where q and s were both equal to 1.

 

PS Do I need a certain number of posts before I can use LaTex? Or is that an option in the toolbar that I just haven't found yet?