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: