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Friday, March 10, 2017

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:



13 comments:

  1. Hi Vijay,

    Amazing Work once again. I am truly astounded by the amount of work and effort you put into this assignment.

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    Replies
    1. Hello Bhavik! Thank you. This is probably because this is less of an assignment and more of a journey for me. Hopefully this will all amount to something.

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  2. Hello Vijay,
    We've made it to the half-way point! You've obviously made a lot of discoveries these past weeks, and I was wondering if you'd been giving any thought to how you'd go about organizing/summarizing your research come May.

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    Replies
    1. Hello Nicole! I was thinking that my senior project would be a math paper that I write, in which case each blog post mostly fits. If I prove anything significant, then I will focus my presentation on the proof for that. If I don't then I will focus more on the many things I tried.

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  3. Hi Vijay. It seems like you are putting a lot of effort into this research. Once again, the calculations were presented clearly and easy to understand. Like the previous weeks, I really enjoyed the calculator at the end. I'm looking forward to next week's blog!

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    Replies
    1. Hello Jack! I'm happy to know that everything made sense. I must admit that this week's post was similar to last week's. Hopefully I will have something entirely new for next time.

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  4. Hey Vijay! I am excited to see how far you have progressed in this research project. The proofs were easy to follow, and like always, I enjoy the addition of the calculator at the end. I look forward to seeing and attending your in-person final presentation!

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    Replies
    1. Hello Gokul! You have dispelled my worries about my calculators being too "obscure." I was worried that the calculators were used to calculate something meaningless to the average reader. I guess the post before each week's calculator did a good job of explaining the significance of the calculator.

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  5. HI Vijay! I originally found this project quite interesting, but thanks to your in depth explanations and meticulous work I am now enthralled by what you are finding. Also, I enjoy all of the calculators at the ends. I cannot wait to find out more of what you discover next week!

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    Replies
    1. Hello Jackson! I am glad to know that I have shown another side of mathematics to you. I also love learning more about different mathematical topics. If you ever want to learn more about different topics in mathematics, you could visit the Numberphile channel on YouTube.

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  6. Hi Vijay! Your research continues to amaze me. What are you planning on working on for the second half of the project?

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    Replies
    1. Hello Carla! I plan on examining the different relations and how they could combine to make a 3x3 magic square of squares. I also plan on writing a math paper near the end of my project. I also might consider easier variations of the magic square of squares problem.

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  7. Hi Vijay! I am so impressed with all of the work you have done! It's going to take me a while to understand it all, but the work you have done and the progress you have made is nothing short of incredible!

    ReplyDelete