Week 1

     If there is one thing I learned this week, I learned that there is an immense amount of research on the 3x3 magic square of squares. From ways of generating magic squares to results of computer-run searches to seemingly arbitrary limitations on what certain numbers in the magic square of squares can be. I'll explain a very small but very interesting portion of this research.

     In any magic square,

$$A\ B\ C\\D\ E\ F\\G\ H\ I$$

we know that the rows add up to the same number,

$A+B+C=S$

$D+E+F=S$

$G+H+I=S$

the columns add up to that same number

$A+D+G=S$

$B+E+H=S$

$C+F+I=S$

as well as the diagonals

$A+E+I=S$

$C+E+G=S$

     From this we can say that

$$\left(A+E+I\right) + \left(D+E+F\right) + \left(C+E+G\right)=\left(A+D+G\right) + \left(C+F+I\right)+S$$

By rearranging the equation we get

$$A+E+I+D+E+F+C+E+G-A-D-G-C-F-I=S=3E$$

this means

$$3E=S$$

that the magic sum is 3 times the center number

We also know that

$$\left(A+B+C\right) + \left(D+E+F\right) + \left(G+H+I\right)=\left(A+E+I\right) + \left(D+E+F\right) + \left(C+E+G\right)$$

By rearranging this we get

$$A+B+C+D+E+F+G+H+I-A-E-I-D-E-F-C-E-G=0=B+H-2E$$

this means

$$2E=B+H$$

By similar proofs we also know that

$2E=A+I$
$2E=C+G$
$2E=D+F$

For a magic square of squares, each of these capital letters are perfect squares so let's say $A=a^2, B=b^2...$
This means that we need some value $$2e^2=a^2+i^2=b^2+h^2=c^2+g^2=f^2+d^2$$

Can you think of any number that is a sum of squares 4 different ways? It seems very... elusive.

Luckily, we have a way of generating these:
$$\left(X^2+Y^2\right)\left(A^2+B^2\right)=\left(XB\pm YA\right)^2+\left(XA \mp YB\right)^2$$
By picking values for $X, Y, A$ and $B$, we can generate two sum of squares. We can then plug in the various combinations of these into the left side and repeat the process. I'll let you try this out at the bottom of the page.

For now, here's my abstract:

Have you heard of a magic square? A magic square is an array of numbers in which the columns, rows and diagonals add up to the same number. Mathematicians have wondered if one could make a “3-by-3 magic square of squares,” a 3-by-3 magic square where the number in each box is the square of an integer. This senior project is conducted at ASU and focuses on constructing a 3-by-3 magic square of squares or proving that such a magic square cannot be constructed. These branches correspond with approaches to the problem. It might be possible to prove that a magic square exists by trying to generate one. Or, it might be possible to prove that a magic square of squares cannot exist by “modding out” entries to see what arrangements of entries are not possible.  This is certainly a very difficult problem; mathematicians have worked on this problem for quite a long time and still not solved it. However, bits of progress have been made over time and hopefully this project can solve the problem or add some more progress. In addition to the 3x3 Magic Square of Squares, this project spends time on similar easier problems. By solving this problem, we may use it in future number theory proofs.

pick some starting values for $X,Y,A$ and $B$. Pick a value for $n$, the number of repetitions. This should give a number that can be expressed as a sum of squares about $2^{2^n}$ ways. I don't suggest going above $n=3$ because $n=4$ gives about $2^{2^4}=65536$ ways.

X:

Y:

A:

B:

n: