Week 7

So last week (or the week before that) we learned how to multiply one number by another number

But in all seriousness, we learned that if we have a number of the form $2M^2-1$ where $M$ is an integer, then $2\left(M+K\right)^2-1$ describes it's multiples as long as $K \equiv 0$ or $-2M \bmod 2M^2-1$. By adding $K=Q\left(2M^2-1\right)$ or $K=Q\left(2M^2-1\right)-2M$ we are effectively multiplying the number by $\left(2MQ+1\right)^2-2Q^2$ or $\left(2MQ-1\right)^2-2Q^2$ (Sorry if I changed the name of certain variables).

Let us remind ourselves of the reason we are finding the multiples of $2M^2-1$. We were able to find arrow relations between Plato triples using

$$Plato \rightarrow Plato\\\boxed{\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}\\i_{from}=L-1\\i_{to}=J}$$

We wanted to know if we could change our values of $J$ and $L$ such that $M_{from}$ and $M_{to}$ don't change. For example, when $J=L$, you can add any number to both

$$\frac{1}{1}=\frac{2\left(1\right)^2-1}{2\left(1\right)^2-1}=\frac{2\left(2\right)^2-1}{2\left(2\right)^2-1}\cdots$$

This is the reason why every Plato triple has a right arrow going to the next term in the Stifel sequence. By finding more of these, we might be able to fully describe the arrow relations between two multiples

So we need to multiply two numbers by the same $\left(2MQ+1\right)^2-2Q^2$ or $\left(2MQ-1\right)^2-2Q^2$

If we say $\left(2MQ+1\right)^2-2Q^2=7$, then we are also saying that $Q\left(2QM^2-Q+2M\right)=3$. Since $Q$ and $M$ are integers we see that the only solution is $Q=1, M=1$. (We started with $7$ because multiplying by $1$ doesn't make sense and the numbers between can never divide a number of the form $2M^2-1$)

If we say $\left(2MQ-1\right)^2-2Q^2=7$, then we are also saying that $Q\left(2QM^2-Q-2M\right)=3$. Since $Q$ and $M$ are integers we see that the only solutions are $Q=1, M=2$ and $Q=3, M=1$

If we interpret these solutions, we see that the only numbers we can multiply by $7$ and $1$ and $7$. (One of those solutions was basically a duplicate) We can see that $2\left(1\right)^2-1=1\rightarrow 7=2\left(2\right)^2-1$ and that $2\left(2\right)^2-1=7\rightarrow 49=2\left(5\right)^2-1$

I will continue exploring this with numbers higher than $7$, but here is a cool pattern I found related to $2M^2-1$. Firstly, here is a list of $2M^2-1$ up to $M=30$

1

7

17

31

49

71

97

127

161

199

241

287

337

391

449

511

577

647

721

799

881

967

1057

1151

1249

1351

1457

1567

1681

1799

We start at $1$ and move forward $\sqrt{1}$ to $7$. $7^2=49$, the next perfect square after $1$.

If we take $49$ and move forward $\sqrt{49}$ to $287$, we might notice that $287=7*41$. And it just so happens that $41^2=1681$ is the next perfect square.

I haven't looked at why this is, but I suspect it is a side effect of $2M^2-1=N^2$ being a Pellian equation.

Another interesting pattern occurs when we consider number we can multiply $2M^2-1$ by

If we consider $M=2$, the first 10 numbers we can multiply by are

7.0

23.0

41.0

73.0

103.0

151.0

193.0

257.0

311.0

391.0

If we consider $M=3$, the first 10 numbers we can multiply by are

23.0

47.0

113.0

161.0

271.0

343.0

497.0

593.0

791.0

911.0

You may have noticed that 23 is second in $M=2$ and first in $M=3$. This pattern actually continues, so hopefully I will be able to exploit it and discover more patterns

This week's calculator will be up shortly

r:

s:

Week 6

This week, there is no post because...

I'm on break!

But next week I will continue my adventure!

(for the 3x3 magic square of squares, not cookies)

Week 5

Last week, we looked at ways to find out where arrows go between
$$Plato \rightarrow Plato\\\boxed{\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}\\i_{from}=L-1\\i_{to}=J}$$
$$Pythagoras \rightarrow Pythagoras\\\boxed{\frac{m_{from}}{m_{to}}=\frac{J^2-2}{L^2-2}\\i_{from}=\frac{L-3}{2}\\i_{to}=\frac{J-1}{2}\\J,L\text{ are odd}}$$
$$Plato \rightarrow Pythagoras\\\boxed{\frac{m_{from}}{m_{to}}=\frac{J^2-2}{2L^2-1}\\i_{from}=L-1\\i_{to}=\frac{J-1}{2}\\J\text{ is odd}}$$
$$Pythagoras \rightarrow Plato\\\boxed{\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{L^2-2}\\i_{from}=\frac{L-3}{2}\\i_{to}=J\\L\text{ is odd}}$$

But what does this even mean? We could plug in values for $J$ and $L$, but we can't do this infinitely, so we need to strategically hold certain values constant.

Let's hold $i_{from}$ and $i_{to}$ constant. For example, let's see what arrows go from $i_{from}=1$ and $i_{to}=3$ for $Plato \rightarrow Plato$

$1=i_{from}=L-1 \Rightarrow L=2$
$3=i_{to}=J \Rightarrow J=3$

$\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}=\frac{17}{7}$
This means that for this example that
$\left(m_{from},m_{to}\right) = \left(17,7\right),\left(34,14\right)\cdots$

However, this is not always so simple. Consider the arrows from $m_{from}=8$ to $m_{to}=14$

$8=i_{from}=L-1 \Rightarrow L=9$
$14=i_{to}=J \Rightarrow J=14$

$\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}=\frac{391}{161}$
So that means

$\left(m_{from},m_{to}\right) = \left(391,161\right),\left(782,322\right)\cdots$
Right?

Actually, no.
$\frac{391}{161}$ can be simplified to $\frac{17}{7}$ and gives

$\left(m_{from},m_{to}\right) = \left(17,7\right),\left(34,14\right)\cdots$

So it turns out that
$$p\text{ is a multiple of } \frac{2J^2-1}{gcf\left(2J^2-1,2L^2-1\right)}$$

$$q\text{ is a multiple of } \frac{2L^2-1}{gcf\left(2J^2-1,2L^2-1\right)}$$

Dividing by the greatest common factor is a way of showing that we simplified the numerator and denominator. To find the GCF, we could use the Euclidean algorithm (which last week's calculator uses), but we cannot calculate this for infinitely numbers. We could also use $\bmod$ to see what we can find.

Let's says that $2M^2-1$ is divisible by $n$
$2M^2-1 \equiv 0 \bmod n$
If we add $K$ to $M$ we get
$2\left(M+K\right)^2-1 \equiv 0 \bmod n$
We know that the difference between these two is $0 \bmod n$
$2\left(2MK+K^2\right) = 2\left(K\right)\left(2M+K\right) \equiv 0 \bmod n$
Since $2M^2-1$ is odd and $n$ divides $2M^2-1$, $n$ is odd
this means $2$ is never $0 \bmod n$ and either
$K \equiv 0 \bmod n$ or $K \equiv -2M \bmod n$

This means that if $2J^2-1$ and $2L^2-1$ are both divisible by a number $n$, then they differ by either $0 \bmod n$, $2J \bmod n$, or $2L \bmod n$

This is as far as I got with holding $i_{from}$ and $i_{to}$ constant


Now we will consider holding $i_{from}$ and $m_{from}$ constant

From $\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}$, we get
$m_{from}\left(2L^2-1\right)=m_{to}\left(2J^2-1\right)$
This means that there are a finite number of arrows emanating from a triple
The number of arrows is the number of divisors $m_{from}\left(2L^2-1\right)$ has that are of
the form $2J^2-1$

This seemed somewhat difficult, so I set it aside for now


Next we will hold $m_{from}$ and $m_{to}$ constant

If we have $\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}$ and found values that work for it,
then we should be able to multiply $2J^2-1$ and $2L^2-1$ by the same number and keep
$m_{from}$ and $m_{to}$ constant.

Earlier we said that if $2M^2-1$ is divisible by $n$, then adding $K$ to $M$ would result in a number also divisible by $n$ as long as $K \equiv 0$ or $-2M \bmod n$

What if $n=2M^2-1$?

$2M^2-1$ divides $2M^2-1$ and if we let $K = Q\left(2M^2-1\right)$ or $Q\left(2M^2-1\right)-2M$ then we find multiples of $2M^2-1$. We essentially multiplied $2M^2-1$ by a number.

What number is this? If we substitute our values of $K$ into $2\left(M+K\right)^2-1$ and divide by $2M^2-1$, we get
$$\left(2QM+1\right)^2-2Q^2$$
or
$$\left(2QM-1\right)^2-2Q^2$$

So if we want to multiply $2J^2-1$ and $2L^2-1$ by the same number, we must ask when
$\left(2QM+1\right)^2-2Q^2=\left(2Q'M'+1\right)^2-2Q'^2$

This is as far as I got, but I do plan on looking at the previous equation as Pellian to find solutions (note that $\left(2QM+1\right)^2-2Q^2$ kinda looks like $X^2-2Y^2$)

Next week will be my Spring Break

I will soon be working a lot with Pellian equations, and I might use Brahmagupta's identity to compose Pellian equations together. Pellian equations are equations of the form
$$X^2-NY^2=k$$

Brahmagupta discovered that if you have
$X_1^2-NY_1^2=k_1$
$X_2^2-NY_2^2=k_2$
then the Pellian equations
$\left(X_1X_2+NY_1Y_2\right)^2-N\left(X_1Y_2+X_2Y_1\right)^2=k_1k_2$
$\left(X_1X_2-NY_1Y_2\right)^2-N\left(X_1Y_2-X_2Y_1\right)^2=k_1k_2$
are true

This calculator will allow you to compose two Pellian equations together

X1:

Y1:

X2:

Y2:

N:

Week 4

Last week on "Don't be a Square: The Magic Square of Squares"...

Our brave adventurer made strange discoveries,

new friends.

and a lot of progress.

This time, Vijay continues on and explores non-primitive pythagorean triples.

---

Non-primitive pythagorean triples $X^2+Y^2=Z^2$ are pythagorean triples where $X$ and $Y$ and not relatively prime. In other words, they can be "reduced."

For example, the pythagorean triple $6^2+8^2=10^2$ is non-primitve because it can be reduced to $3^2+4^2=5^2$

If we include non-primitive triples, then we must also consider Fermat family triples. We eliminated them last time because $Y-X=1$ for Fermat triples, but this is not the case for non-primitive triples.

Last time, we learned about the white arrow and that it meant the that $X+Y$ of one triple equals $Y-X$ of another triple.

Here are all the possible cases for the arrow relation:

$p \times Fermat \rightarrow q \times Fermat$
$p \times Fermat \rightarrow q \times Plato$
$p \times Fermat \rightarrow q \times Pythagoras$
$p \times Plato \rightarrow q \times Fermat$
$p \times Plato \rightarrow q \times Plato$
$p \times Plato \rightarrow q \times Pythagoras$
$p \times Pythagoras \rightarrow q \times Fermat$
$p \times Pythagora \rightarrow q \times Plato$
$p \times Pythagoras \rightarrow q \times Pythagoras$

We need to find out when each of these are possible. Here are the mathematical expressions for the sums and difference of the primitive type of each triangle

$X+Y$ for Fermat:
$\frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right)$

$Y-X$ for Fermat:
$1$

$X+Y$ for Plato:
$2n^2+4n+1$

$Y-X$ for Plato:
$2n^2-1$

$X+Y$ for Pythagoras:
$4m^2+12m+7$

$Y-X$ for Pythagoras:
$4m^2+4m-1$

This means that our list of cases for the arrow relation becomes

$p \times \frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right) = q \times 1$
$p \times \frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right) = q \times 2n^2-1$
$p \times \frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right) = q \times 4m^2+4m-1$
$p \times 2n^2+4n+1 = q \times 1$
$p \times 2n^2+4n+1 = q \times 2n^2-1$
$p \times 2n^2+4n+1 = q \times 4m^2+4m-1$
$p \times 4m^2+12m+7 = q \times 1$
$p \times 4m^2+12m+7 = q \times 2n^2-1$
$p \times 4m^2+12m+7 = q \times 4m^2+4m-1$

Because $X+Y$ and $Y-X$ are $1 \bmod 2$ for a pythagorean triple, modding all these equations by $2$ gives

$p \times 1 \bmod 2 = q \times 1 \bmod 2$
$p \bmod 2 = q \bmod 2$

This means arrows can only be drawn between two odd multiple triples or between two even multiple triples. This also happens to be true for the red, blue and green lines, so we can actually ignore even multiples since a reduced version can be found.

This is something I noticed very recently, but really it's ... something I should have noticed sooner

Anyways, let's look at some of these cases.

$$p \times 2n_1^2+4n_1+1 = q \times 2n_2^2-1$$

$$\frac{p}{q} \times 2n_1^2+4n_1+1 =  2n_2^2-1$$

$$\frac{p}{q} \times n_1^2+2n_1+\frac{1}{2} =  n_2^2-\frac{1}{2}$$

Let $n_2=n_1+k$

$$\frac{p}{q} \times n_1^2+2n_1+\frac{1}{2} =  n_1^2+2n_1k+k^2-\frac{1}{2}$$

$$0=n_1^2+2n_1k+k^2-\frac{1}{2}-\frac{p}{q}n_1^2-\frac{p}{q}2n_1-\frac{p}{q}\frac{1}{2}$$

By the quadratic formula, we get 

$$k=\frac{-2n_1 \pm \sqrt{\left(2n_1\right)^2-4\left(n_1^2-\frac{1}{2}-\frac{p}{q}n_1^2-\frac{p}{q}2n_1-\frac{p}{q}\frac{1}{2}\right)}}{2}$$

$$k=-n_1 \pm \sqrt{\frac{1}{2}+\frac{p}{q}n_1^2+\frac{p}{q}2n_1+\frac{p}{q}\frac{1}{2}}$$

We need

$$\frac{1}{2}+\frac{p}{q}n_1^2+\frac{p}{q}2n_1+\frac{p}{q}\frac{1}{2}=J^2$$

$$\frac{p}{q}n_1^2+\frac{p}{q}2n_1+\frac{p}{q}\frac{1}{2}+\frac{1}{2}-J^2=0$$

By the quadratic equation

$$n_1=\frac{-\frac{p}{q}2 \pm \sqrt{\left(\frac{p}{q}2\right)^2-4\left(\frac{p}{q}\right)\left(\frac{p}{q}\frac{1}{2}+\frac{1}{2}-J^2\right)}}{2\frac{p}{q}}$$

$$n_1=-1 \pm \sqrt{\frac{1}{2}-\frac{1}{2}\frac{q}{p}+\frac{q}{p}J^2}$$

We need

$$\frac{1}{2}-\frac{1}{2}\frac{q}{p}+\frac{q}{p}J^2=L^2$$

$$\frac{q}{p}\left(-\frac{1}{2}+J^2\right)=L^2-\frac{1}{2}$$

$$\frac{p}{q}=\frac{J^2-\frac{1}{2}}{L^2-\frac{1}{2}}$$

To summarize this means

$$\frac{p}{q}=\frac{J^2-\frac{1}{2}}{L^2-\frac{1}{2}}=\frac{2J^2-1}{2L^2-1}$$

$$n_1=L-1$$

$$n_2=J$$

By a similar proofs, we can get equations for three more cases. Here I will rewrite all four of them with clearer variables. $m$ will indicate the multiple of the primitive series while $i$ will indicate the index of the triple in the series

$Plato \rightarrow Plato$

$J$ and $L$ are integers

$$\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}$$

$$i_{from}=L-1$$

$$i_{to}=J$$

$Pythagoras \rightarrow Pythagoras$

$J$ and $L$ are odd integers

$$\frac{m_{from}}{m_{to}}=\frac{J^2-2}{L^2-2}$$

$$i_{from}=\frac{L-3}{2}$$

$$i_{to}=\frac{J-1}{2}$$

$Pythagoras \rightarrow Plato$

$J$ and $L$ are integers and $L$ is odd

$$\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{L^2-2}$$

$$i_{from}=\frac{L-3}{2}$$

$$i_{to}=J$$

$Plato \rightarrow Pythagoras$

$J$ and $L$ are integers and $J$ is odd

$$\frac{m_{from}}{m_{to}}=\frac{J^2-2}{2L^2-1}$$

$$i_{from}=L-1$$

$$i_{to}=\frac{J-1}{2}$$

If we look a bit closer at

$$\frac{m_{from}}{m_{to}}=\frac{J^2-2}{2L^2-1}$$

$$m_{from}\left(2L^2-1\right)=\left(J^2-2\right)m_{to}$$

Since $J^2$ is an odd square, we can say

$$m_{from}\left(2L^2-1\right)=\left(8\left(\frac{n^2+n}{2}\right)+1-2\right)m_{to}$$

When we mod out by $8$ we get

$$m_{from}\left(2L^2-1\right)=\left(7 \bmod 8\right)m_{to}$$

This chart shows that if $L$ is odd, then $2L^2-1=1 \bmod 8$ and

if $L$ is even, then $2L^2-1=7 \bmod 8$

so if $L$ is odd, we get

$$m_{from}\left(1\bmod8\right)=\left(7 \bmod 8\right)m_{to}$$

$$m_{from} \bmod 8 = -m_{to} \bmod 8$$

if $L$ is even, we get

$$m_{from}\left(7\bmod8\right)=\left(7 \bmod 8\right)m_{to}$$

$$m_{from} \bmod 8 = m_{to} \bmod 8$$

I have not yet tried this out on the other cases.


I have noticed that the arrows between the two families occur in pairs. I will explain more in this week's calculator.


If we again look at
$$m_{from}\left(2L^2-1\right)=\left(J^2-2\right)m_{to}$$

and let $m_{from}=1$ and $m_{to}=1$ then we get the Pellian equation

$$J^2-2L^2=1$$

which can be used to find value of $J$ and $L$ for $m_{from}=1$ and $m_{to}=1$

When an arrow goes to a Fermat family triple, it actually goes to all Fermat family triples of that multiple since they all share the $Y-X$ term.


That leaves only

$p \times \frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right) = q \times 2n^2-1$

$p \times \frac{1}{2}\left(\left(1+\sqrt{2}\right)^{2k+1}+\left(1-\sqrt{2}\right)^{2k+1}\right) = q \times 4m^2+4m-1$

which I have not gotten to yet.


After I examine these two, I will start examining the red, green, and blue line relations.

And that is where I am at.

Here is a calculator that finds where arrows go between according to the equations in this post. The first line tells you the triples in terms of family(index,multiple) while the second line tells you the explicit values of the triples.

To see the arrow pairs, find values of $J$ and $L$ that work for Pythagoras => Plato. Then switch the values of $J$ and $L$ and hit Plato => Pythagoras. You should find a pattern like the picture below. This can also be observed with arrows within the same family. find values for $J$ and $L$ where $J \neq L$ that work for Plato => Plato. and then switch the two values and again try Plato => Plato. The same again works for Pythagoras => Pythagoras.

J:

L:

Week 3

     Last week, I worked on finding ways to express the 3x3 magic square of squares and this week, I found many more ways. Let's pick up where we left off.

We were looking for
$$\left(n_Y+1\ \ \ \ +\ \ \ \ n_Y+2\ \ \ \ ...\ \ \ \ n_Z \right)=\left(n_X+1\ \ \ \ +\ \ \ \ n_X+2\ \ \ \ ...\ \ \ \ n_Y \right)$$
so I thought of this as 
$$\left(m\ \ \ \ +\ \ \ \ m+1\ \ \ \ ...\ \ \ \ n \right)=\left(n+1\ \ \ \ +\ \ \ \ n+2\ \ \ \ ...\ \ \ \ n+k \right)$$
We need to find $m,n,k$ such that
$$\frac{\left( n+m \right) \left( n-m+1 \right) }{2}=\frac{\left( 2n+k+1 \right) \left( k \right) }{2}$$
$$n^2-m^2+n+m=\left(2n+k+1\right)\left(k\right)$$
$$n^2-m^2-n+m=\left(2n+k+1\right)\left(k\right)-2n$$
$$\left(n-m\right)\left(n+m-1\right)=\left(2n+k+1\right)\left(k\right)-2n$$
Let $w=n-m$ and $x=n+m+1$. Also note that $2n=w+x+1$
$$wx=\left(w+x+k+2\right)\left(k\right)-\left(w+x+1\right)$$

$$0=k^2+wk+xk+2k-wx-w-x-1$$
$$0=k^2+k\left(w+x+1\right)-\left(w+1\right)\left(x+1\right)$$
Through the quadratic formula, we get
$$k=\frac{-w-x-2 \pm \sqrt{\left(x+x+2\right)^2 -4\left(-\left(w+1\right)\left(x+1\right)\right)}}{2}$$
Essentially we need the $b^2-4ac$ term to be a perfect square, but first we'll consider $b^2$ and $b^2+4ac$
$$b^2+4ac=\left(w+x+2\right)^2 +4\left(-\left(w+1\right)\left(x+1\right)\right)=w^2-2wx+x^2=\left(x-w\right)^2$$
So $b^2+4ac$ is a perfect square.
$b^2$ is also a perfect square.
This means that if $b^2 -4ac$ is a perfect square, it is the last term in an arithmetic progression of squares which is what we were looking for in the first place.

This seems like all the ways of expressing a magic square I would find, so I decided to take a step back and try a different approach. Besides, I was getting really frustrated anyways

We need the 8 sums of the magic square to be equal.
$A+B+C=S$
$D+E+F=S$

$G+H+I=S$

$A+D+G=S$

$B+E+H=S$

$C+F+I=S$

$A+E+I=S$

$C+E+G=S$

If the following are true:
$B+H=D+F=2E$
$B+D=2I$
$B+F=2G$
$H+D=2C$
$H+F=2A$

then
$A+B+C=\frac{H+F}{2}+\left(D+F-H\right)+\frac{H+D}{2}=\frac{3F+3D}{2}$
$A+D+G=\frac{H+F}{2}+D+\frac{B+F}{2}=\frac{2D+B+H+2F}{2}=\frac{3F+3D}{2}$
$D+E+F=\frac{2D+2E+2F}{2}=\frac{3D+3F}{2}$
$A+E+I=\frac{H+F}{2}+\frac{D+F}{2}+\frac{B+D}{2}=\frac{2D+B+H+2F}{2}=\frac{3F+3D}{2}$
This continues for all 8 sums we need.

In squares where $B+H=D+F=2E$, we can say $B<D<F<H$ without loss of generality

(By "without loss of generality," I mean a square with $B+H=D+F=2E$ but not $B<D<F<H$ can just be rotated or reflected so that $B<D<F<H$)

We can express the magic square of squares as 6 arithmetic sequences of squares:
$b^2,e^2,h^2$
$d^2,e^2,f^2$
$b^2,i^2,d^2$
$b^2,g^2,f^2$
$d^2,c^2,h^2$
$f^2,a^2,h^2$

Here's my best attempt to visually represent this

Each rectangle is an arithmetic progression of size 3 and each letter is an entry. Let's also give each progression a name.

Each rectangle can be related to another rectangle in a couple of ways

1. First terms match
2. Middle terms match
3. Last terms match
4. Last term of one matches first term of other

If we draw out these relationships, we get

Where the red line mean the first terms match, green lines mean the middle terms match, blue lines mean the last terms match. and white arrows go from an arithmetic progression whose last term matches the first term of the arithmetic progression it goes to.

An arithmetic progression $r^2,s^2,t^2$ can also be expressed as a Pythagorean triple $X^2+Y^2=Z^2$ through

$r=Y-X$

$s=Z$

$t=X+Y$

For the white arrow this means that $X_1+Y_1=Y_2-X_2$

which we can find in Stifel and Ozanam sequence.

A Stifel sequence generates all primitive triples in the Plato family

$$1\frac13 , 2\frac25 , 3\frac37 ...$$

to get the triples, express them as an improper fraction

$$\frac43 , \frac{12}{5} , \frac{24}{7}...$$

and we get

$3^2+4^2=5^2$, $5^2+12^2=13^2$, $7^2 + 24^2 = 25^2$...

Also notice that $3+4=12-5$, $5+12=24-7$...

This means we would draw white arrows between arithmetic progressions of squares generated by consecutive terms in this progression.

A Ozanam sequence generates all primitve triples in the Pythagoras family

$$1\frac78 , 2\frac{11}{12} , 3\frac{15}{16}...$$

to get the triples, express them as an improper fraction

$$\frac{15}{8} , \frac{35}{12} , \frac{63}{16}...$$

and we get $8^2+15^2=17^2$, $12^2+35^2=37^2$, and $16^2+63^2=65^2$

Again notice that $8+15=35-12$, $12+35=63-16$...

This means we would draw white arrows between arithmetic progressions of squares generated by consecutive terms in this progression.

The primitive triples in the Fermat family have $Y-X=1$ which would yield $1$ as an entry in the magic square of squares. Morgenstern proved that 1 cannot be an entry in a magic square of squares, so we don't need to consider the Fermat family.

So we know to draw arrows between consecutive terms of the Stifel and Ozanam sequences, but should we draw any arrows between them?

We would need

the $X+Y$ of the nth term of Stifel = $Y-X$ of the mth term of Ozanam

$2n^2+4n+1=4m^2+4m-1$

let $n=m+k$

$2m^2+4mk+2k^2+4m+4k+1=4m^2+4m-1$

$0=-2k^2-4mk-4k+2m^2-2$

$0=k^2+k\left(2m+2\right)+\left(1-m^2\right)$

By the quadratic formula, we get

$$\frac{-2m-2 \pm \sqrt{4\left(m+1\right)^2+4\left(m^2-1\right)}}{2}$$

$$-m-1 \pm \sqrt{2\left(m\right)\left(m+1\right)}$$

So we need to prove that $2\left(m\right)\left(m+1\right)$ is a perfect square

By reversing Dickson's method, we can find values for $m$

If there are $u,v,w$ such that $w^2=2uv$, then there is a Pythagorean triple can be made by

$x=w+u$

$y=w+v$

$z=w+u+v$

If we reverse this so that $u=m$ and $v=m+1$, we are looking for a Pythagorean triple where $y-x=1$, or in other words, one from the Fermat family.

So Pythagorean triples from the Fermat family dictate where we can draw arrows from the Stifel sequence to the Ozanam sequence, but what about from the Ozanam sequence from the Stifel sequence?

We would need

$Y-X$ of the nth term of Stifel = $X+Y$ of the mth term of Ozanam

$2n^2-1=4m^2+12m+7$

let $n=m+k$

$2m^2+4mk+2k^2-1=4m^2+12m+7$

$0=-2k^2-4mk+2m^2+12m+8$

$0=k^2+k\left(2m\right)+\left(-6m-m^2-4\right)$

By the quadratic formula,

$$k=\frac{-2m \pm \sqrt{4m^2+4\left(6m+m^2+4\right)}}{2}$$

$$k=-m \pm \sqrt{2\left( m+1\right)\left(m+2\right)}$$

We again use reverse Dickson's method to find that we must find a Pythagorean triple from the Fermat family.

If we visualize Stifel and Ozanam sequences then we get

Our work with arrows between the two sequence would mean that later on we would see multiple of something similar to

So far, I've only considered primitive triples, only considered the white arrow, and only considered Plato and Pythagoras families. So there's a lot to learn.

And that's where I'm currently at.

Here we have a calculator that will let you see the nth triple of the Plato, Pythagoras, or Fermat family. Just enter $n>0$ and select which family you want. For large values of $n$, the Fermat family triples get very large. So large in fact that the programming variables overflow. For $n>20$ it doesn't work properly and for $n>402$ it says "infinity^2 + infinity^2 = infinity^2." Just something I found funny.