# Pythagorean Triples? (Problem)

I’ve spent some time thinking about how I want to present things I love on this blog, and one format I’ve come up with is a Problem/Solution series. The idea here is to have two posts with the same name – in one of them, I will explain an interesting problem and try to give some prodding questions about how an interested reader might play around with it. Then, in a second post, I’ll try to show how a mathematician might look at the problem, and then I’ll provide a solution to the problem.

The goal here is twofold. Firstly, I’d like to give people an opportunity to try things for themselves. Think of this like a puzzle – you don’t have to know much math to give things a try. Full answers often take a very long time to find, but even finding part of the answer can be very gratifying. The second reason I like this format is that I can use it to give anyone reading a “look into the mind” of a mathematician. I’ll build things up one step at a time, hopefully shedding light on important ideas along the way and showing how the math really does have a story behind the scenes. I’d strongly encourage playing with problems on your own before reading my answer. You’ll get a lot more out of it that way, even if it’s just for 5 minutes.

Anyways, on to the first problem! For the math people out there, the “name” of the problem I will be laying out is the classification of all Pythagorean triples. If you don’t know what that means, don’t worry about it. It will be explained.

In a previous post (So What is a Proof?), I showed why the Pythagorean Theorem is true. So, we now know that now. But we can keep asking questions. For example, we now have an equation about right triangles: $a^2+b^2 = c^2.$

Now, like with any other equation, it is natural to try to find some solutions. We can choose any values for a and b that we want, and then ask what value c is. Well, we can do a “square root” on both sides of the previous equation, and now we get $c = \sqrt{a^2+b^2}.$

We might ask what sort of number c is, and we then realize that c is the distance between two points on that triangle, and so we now have learned how to calculate distances! If you are into geometry, you might want to know if there is a version of this for triangles that do not have 90 degree angles (there is, it’s called the Law of Cosines for those who are interested). We might have noticed that we can divide both sides by c squared and rearrange a bit to get a new equation $\bigg( \dfrac{a}{c} \bigg)^2 + \bigg( \dfrac{b}{c} \bigg)^2 = 1.$

Now this looks like a question about fractions. Well, we never actually said that a and b had to be whole numbers, so maybe it isn’t really a fraction… but what if it were? This is the question that comes most naturally to me, because I am fascinated by whole numbers. Might I be able to find some whole numbers that make this equation work?

This isn’t immediately obvious one way or the other. Equations can be rather unpredictable, especially when you want every number to be a whole number. I’ll give an example to show what I mean. One of the questions related to this which people began talking about a long time ago was “What happens if you change the 2 to something else?” So what about an equation like $a^3 + b^3 = c^3$

or $a^{11}+ b^{11} = c^{11}$

or any other value in place of 2… maybe those have some whole-number solutions too? As we will soon see, when we left the exponent as 2, there are plenty of whole-number solutions. What took hundreds upon hundreds of years to discover is that if you change the exponent to 3, or 4, or 5, or any whole number larger than 2, the number of solutions drops to ZERO. (Well, except for if you make a=0 and b=c, say, but that’s pretty boring, and when the exponent is 2 there are some not-so-boring examples.) For those who are interested, this is called Fermat’s Last Theorem, and we have only known this amazing fact since 1996. There are tons of good books and YouTube videos on this topic, I’d highly recommend surfing a few.

The main point of that excursion is just to say that equations don’t always play nice. Weird things happen. Even with the regular equation, most of the time you don’t get whole numbers. But what about this equation? What kinds of answers does it have? Now, I can say why I described the problem the way I did. If we find three positive whole numbers (a,b,c) that make the equation $a^2+b^2=c^2$

true, we call (a,b,c) a Pythagorean triple (since it’s three numbers that satisfy Pythagoras’ Theorem). Another way we could ask this question might be what are all the right triangles where every side is a whole number of units long?

Try it yourself! If you check out the post with the same title with the word “Problem” with “Solution”, if I’ve posted it yet, you’ll be able to see a complete answer, along with all the steps a mathematician might take to get there.

## 2 thoughts on “Pythagorean Triples? (Problem)”

1. Tanner Carawan says:

There’s a missing plus sign

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1. Will Craig says:

I think I found it and correct it, thanks!

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