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Friday, March 31, 2017

Week 8

This week, instead of looking at specific families of Pythagorean triples, I looked at all Pythagorean triples. I'll explain how I got there by starting from the beginning.

So I reviewed this thing:


my visual representation of the 6 arithmetic progressions of squares I needed to find to construct a 3x3 Magic Square of Squares. Maybe I could start out by finding a smaller piece like this:


This would entail finding 3 Pythagorean triples and using our method of converting Pythagorean triples into arithmetic progressions of squares.

If we have a Pythagorean triple \left(a,b,c\right), then our arithmetic progression is \left(b-a\right)^2,a^2+b^2,\left(a+b\right)^2. Note that the difference between the first and last term is 4ab. If we give these triples names, then we can start writing equations.


We know that 4ab=4mn+4xy \Rightarrow ab=mn+xy

We also know that m+n=y-x \Rightarrow y=m+n+x

By substituting, this gives us ab=mn + mx + nx + x^2=\left(m+x\right)\left(n+x\right)


We also know that b-a=n-m \Rightarrow n=b-a+m

By again substituting, we get ab=\left(m+x\right)\left(b-a+m+x\right)

If we let k=m+x, we get ab=k\left(b-a+k\right)

The only possible value for k is a, so we see that m+x=k=a


If we instead rearrange b-a=n-m as m=n+a-b, substitute, and let l=n+x, we get
ab=\left(l+a-b\right)l, giving us that n+x=l=b


We already knew that that b-a=n-m (since the Fermat family had b-a=1, this is sort of a more general version), but we learned that b-n=a-m=x

If I look at all Pythagorean triples, I might be able to divide them up into classes based on their b-a value. So how does one look at all Pythagorean triples?

Well, if we pick two positive integers r and s where r>s, then we can generate a Pythagorean triple \left(r^2-s^2,2rs,r^2+s^2\right). If we put r and s on the axes of an Excel chart, and write out the legs of each triangle, we get

Note that the grayed-out boxes contain non-primitive triples and that the upper right corner is empty because r>s and that this goes only r<15

Since we want to divide these up into classes based on b-a, let's display that instead


Now, let's highlight Fermat family triples



If we look at the \left(r,s\right) values for Fermat family triples we get
\left(2,1\right)
\left(5,2\right)
\left(12,5\right)

By now you may have noticed some patterns


\left(2,1\right)
\left(2*2+1,2\right)=\left(5,2\right)
\left(5*2+2,5\right)=\left(12,5\right)

And more importantly, this pattern holds for other numbers (although there seem to be 2 \left(r,s\right) starting points needed)

So we have divided up Pythagorean triples into different classes. Using this will make it much easier to find b-n=a-m=x

This week's calculator will be up shortly