Week 9

Last week, we divided up all primitive Pythagorean triples into classes based on the value $b-a$ where $b$ is the long leg of triangle and $a$ is the short leg of the triangle. What we learned from this is that triangles of a certain class start at $\left(r,s\right)$ and keep extending onto $\left(2r+s,r\right)$, $\left(2\left(2r+s\right)+r,2r+s\right)$.... (This is actually related to a recurrence relation called Pell numbers)

I wondered if we could locate the first triangle from each class, so I asked myself when there is nothing before an $\left(r,s\right)$. The term before $\left(r,s\right)$ would be $\left(s,r-2s\right)$. This is not a valid value when at least one of the parts is less than $1$ or when the first half is less than the second half

If we look at this first condition, we see that either $s$ or $r-2s$ is less than $0$. If we are given that $\left(r,s\right)$ is valid then we know that $s$ is greater than $0$. This only leaves the equation $r-2s<0$ or $r<2s$

The other condition gives the equation $s<r-2s$ or $r>3s$

We see that both of these equations seem to describe lines on a graph and if we look back to our r vs. s spreadsheet, we see

This means that everything above the blue line and everything below the red line is the first of its class in that method of finding the next triple of the same class. You can also see that the triples between the two lines already had a smaller triple of the same class outside of the lines.

This actually helped cut down on calculation time for a calculator I made, but to explain that I first need to explain something else.

Let's say we draw out a Pythagorean triple as a triangle

We could subtract some $k$ from both $b$ and $a$ and not change its class since $b-a=\left(b-k\right)-\left(a-k\right)$ (although this new triangle might not represent a Pythagorean triple) 

And we could do this for any value of $k$, so we know triangles of the same class would have their hypotenuse from the origin to anywhere along this line.

Remember that this is for triangle with the same value for $b-a$. If we instead look at triangles with the same value for $b+a$, we get

This is important because if we review our arithmetic sequences

we see that there are many triangles that share their $b-a$ or $b+a$ class with another triangle. If we try drawing this out like we did earlier, we get

To clean this up, we will only show the endpoint of each hypotenuse

This includes almost all the requirements for a 3x3 magic square of squares, but we will worry about the last bit later.

Now, I will describe my program.

First, it collected the first thousand triples of a certain $b-a$ class

Next, for every pair of triples, it essentially took the two triples

and used them to calculate the third triple

And so I got solutions to some classes. For example, for the class $23$, I got the three triples $\left(119,408,425\right)$, $\left(275,252,373\right)$, and $\left(156,133,205\right)$

These give us $289^2, 425^2, 527^2$,  $23^2, 373^2, 527^2$, and $23^2, 205^2, 289^2$

We notice that we can take this

and fill in the values

So we have found out how to generate this part of a magic square of squares, but we have still not yet figured out the other parts of the magic square of squares

This week's calculator will be up shortly