Week 10

Last week, I used a program to try finding this structure

To recap, my program essentially generated a pair of triangles in the same $b-a$ class

and generated a third triangle

If the full structure exists, one would expect my program to return at least two results for one of the classes. This is because for a certain class, my program would give the three red triangles once and the three green triangle once.

However, I only got one result for each class. This could mean that this overall structure is so large that the variables in my program overflowed before it reached. One thing I realized about my program is that it checks only primitive triangles meaning that I accidentally skipped some triangles in each $b-a$ class.

Last week, I also examined an excel spreadsheet of the $b-a$ classes of primitive Pythagorean triples. However, I also examined the $b+a$ classes as well

Each red value indicates a duplicate value. This is important because in our diagram, we have three triangles that belong to the same $a+b$ class.

By looking at this spreadsheet some patterns become immediately clear: duplicate values occur at regular intervals in lines. If we pick a line through the origin with slope $-\frac{r}{s}$, and pick a point it hits, we should be able to draw a line of slope $\frac{r-s}{r+s}$ through this point it hits and find duplicate values. For example, if we pick slope $3$, the points we hit are highlighted in blue. If we then draw lines of slope $\frac{3-1}{3+1}=\frac24=\frac12$, we hit duplicate values.

While this is very interesting, it is not what I am focusing on most.

If we return to the structure of Pythagorean triangles

we notice that the white lines showing the class of each triangle converge to four points at the bottom. This means that from these four numbers, we can generate this whole structure. First let's name these points

We need each triangle's hypotenuse to be an integer, so we need $\sqrt{height^2+length^2}=\sqrt{\left(\frac{y-x}{2}\right)^2+\left(\frac{x+y}{2}\right)^2}=\sqrt{\frac{x^2+y^2}{2}}$ to be an integer where $x$ and $y$ are any pair of numbers from $a,b,c$, or $d$. The six possible pairs each correspond to a triangle.

How would we start finding such numbers?

If we start out with

$a_1^2+b^2=2j^2$

$a_2^2+c^2=2k^2$

$a_3^2+d^2=2l^2$

we can then say

$\left(a_1a_2a_3\right)^2+\left(ba_2a_3\right)^2=2\left(ja_2a_3\right)^2$
$\left(a_2a_1a_3\right)^2+\left(ca_1a_3\right)^2=2\left(ka_1a_3\right)^2$
$\left(a_3a_1a_2\right)^2+\left(da_1a_2\right)^2=2\left(la_1a_2\right)^2$

So we now have a number that satisfies our equation three ways

Now we need the pairs of these other three numbers to satisfy the equations

$\left(ba_2a_3\right)^2+\left(ca_1a_3\right)^2=2p^2$

$\left(ba_2a_3\right)^2+\left(da_1a_2\right)^2=2q^2$

$\left(da_1a_2\right)^2+\left(ca_1a_3\right)^2=2r^2$

You might notice that we can simplify these equations to

$\left(ba_2\right)^2+\left(ca_1\right)^2=2p'^2$
$\left(ba_3\right)^2+\left(da_1\right)^2=2q'^2$

$\left(da_2\right)^2+\left(ca_3\right)^2=2r'^2$

If we look at the first equation and our starting equations, we see

$a_1^2+b^2=2j^2$

$a_2^2+c^2=2k^2$

$\left(ba_2\right)^2+\left(ca_1\right)^2=2p'^2$

The first two equations describe Pythagorean triples and the third equation describes a Pythagorean triple made of the previous two Pythagorean triples. So it is possible to "multiply" two Pythagorean triples together?

If we look back to

then we can see that

$23^2+289^2=2*205^2$

$527^2+289^2=2*425^2$

can be multiplied to get

$\left(289*23\right)^2+\left(289*527\right)^2=\left(373*289\right)^2$

this week's calculator will be up shortly