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
