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 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
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:
Y1:
X2:
Y2:
N: