Pythagorean triples and parabolas
I am going to have to give a shout out to Pat’s Blog again in this post, as he was the inspiration. Seriously, if you do not have his blog bookmarked in your feed reader, you need to. He is a fabulous blogger on math topics.
The specific post I that inspired this extension on quadratics is his post on Pythagorean Parabolas. I am not going to duplicate what he said, but show some connections that I found truly interesting.
First off, let’s chat about the really simple way to build Pythagorean Triples he mentions. Honestly, I had not heard of this method before. (I also have to admit, I taught everything except geometry in HS, so that is probably why).
Take any odd number. Say, 3 for example. Square it, so you get 9, divide by 2, so you get 4.5. Now, do the floor and the ceiling on 4.5, and you get 4, and 5. The Pythagorean Triple is 3-4-5. There is an additional interesting thing with this series of triples, and that is because we do the floor and ceiling, the hypotenuse and the other long side will always have a difference of 1.
If we graph the two short sides, and ignore the hypotenuse, so in essence, the short side is the x value, and the long non-hypotenuse is the y value of a triangle, we get the following graph:
Looks a lot like a parabola! In fact, it is. It has an equation of:
\[ f(x)=\frac{1}{2}x^2 -\frac{1}{2} \]
Which makes sense, given the definition of a parabola using the focus and directrix. Pat puts it this way:
If we think about the definition of a parabola as the set of points equally distant from a focus and directrix, we realize the line of the directrix must be the line y = -1 so that, for instance the point (3,4) which is 5 units from the origin/focus will also be 5 units away from the directrix.
So if we multiply all the original triples by 2 (so our first triple is 6-8-10) we get a triple that has a difference of 2 between the two longest sides, so the equation should be:
\[ f(x)=\frac{1}{4}x^2 -1 \]
and sure enough!
But, there are other ways to create Pythagorean triples.
Multiplying the primitive Pythagorean Triples by 3, gives us an equation of \(f(x)=\frac{1}{6}x^{2}-\frac{3}{2}.\)
Does this always work for every Pythagorean Triple though? That is a very interesting question, because we used a very simple, primitive way to create the triples, but there are other ways.
For example, there is this way to create the triples. Take any two numbers, such that \(m<n.\)
\(n^2-m^2, 2mn, n^2+m^2\) .
So, for example, we can use consecutive Fibonacci numbers (for fun, why not), we get the following Pythagorean Triples.
3-4-5! Yay, that matches what we got before. But then it explodes.
5-12-13 , again, a simple triple, found above.
16-30-34, this is not a multiple of a primitive triple as above. but it is a multiple of 8-15-17. The 8-15 is on the ‘times 2’ parabola, while the 16-30 is on a ‘time 4’ parabola, \(f(x)=\frac{1}{8}x^{2}-2\).
39-80-89, This Triple is the first one that is different! It IS a primitive Triple. So, does it follow the pattern found above? Is it a “difference of 9” parabola? Yes! It is! \(f(x)=\frac{1}{18}x^{2}-\frac{9}{2}\)
It seems the generalization of the equation is: \(f_n(x)=\frac{1}{2n}x^2-\frac{n}{2}\) where n is the difference between the hypotenuse and the next longest side, AS LONG AS IT IS A PRIMITIVE TRIPLE!
Let’s test it on this next triple using Fibonacci numbers.
105-208-233 ; This is a primitive Pythagorean Triple, as the 3 numbers are co-prime. The difference between the last two numbers is 25. So, the function that would create the parabola this triple should be: \(f_{25}(x)=\frac{1}{50}x^2-\frac{25}{2}.\)
Yes!
It works. Also, if you substitute 105 into the \(f_{25}(x)\) function, you get an output of 208.
Which begs the question of Why?
I don’t have that answer, nor do I have a proof. Right now, I have a conjecture that for every primitive Pythagorean Triple, \((a,b,c)\) such that \(a^2+b^2=c^2\), and where \(a<b<c\), if \(n=c-b\) then the point \((a,b)\) will be on the parabola described by the function \(f_n(x)=\frac{1}{2n}x^2-\frac{n}{2}\) .
Edit: 24 October
Proof of my conjecture:
Shoutout to TrigOrTreat over on Mastodon who pointed out that if I take my statement that \(n=c-b\) and reframe it as \(c=n+b\) and insert it back into the Pythagorean Theorem and solve for b, the conjectured equation for the they parabola is an outcome.
\[ a^2 + b^2 = (n+b)^2 \] \[ a^2 + b^2 = n^2 + 2nb + b^2 \] \[ 2nb=a^2 -n^2 \] \[ b = \frac{a^2}{2n} - \frac{n}{2} \] Not a hard proof of the conjecture at all.
Pythagorean Triples are really interesting things, and the connections to quadratics is also fascinating!
Edit: 21 October
As I was exploring how to prove this, I came across this interesting video by 3Blue1Brown on calculating triples from complex numbers. Um, WOW! Every Pythagorean Triple can be thought of as the square of any complex number with integer components. (definitely watch the video. It is 14 min well spent)
So, for example, if you take the complex number \((9+12i)\) and square it, the magnitude of this new complex number is \(15\)! And the Pythagorean Triple is (9,12,15).
When you do a whole bunch of these, and rotate the axes to show the new lattice points of the coordinate system, it looks like this:
This does not answer the question of proving the general equation above, but it does demonstrate the fact that the equations for these are quadratics. Still have a way to go in my proof, but this is a really interesting connection between complex numbers and Pythagorean Triples.
It also shows why one formula for finding Pythagorean Triples is the one given above. It is the way to find the new magnitude of the hypotenuse of the two vectors in imaginary space. The video does a wonderful job showing it and demonstrating the proof.