Pythagorean theorem reflection

Life Without Euclid

This post has nothing to do with geometry. I guess I can’t say that exactly (because of the possible geometric representations), but I am not dealing directly with these. I am always intrigued when I think like I want my students to think. It is these moments that keep me going into the classroom hoping for new understandings. There have been times this year where students have made connections that I never have. These innocent realizations are mathematics manifested in its purest form. A similar experience happened to me this morning.

I had been mulling over a problem posed by the NCTM about Pythagorean Triples. I am very familiar with the theory around these special sets of numbers, but have fallen slightly rusty over the years since the number theory classroom. The problem was:

Can you have a Pythagorean Triple without at least one of the entries being even?

This would have been a rather dull problem for me to explore if it wasn’t for that rust which has accrued over time. It allowed me to approach the problem with blessed ignorance. The first approach was to try and remember the formula that generates all triples; this failed miserably. I was sure that one of the stipulations was one of the terms must be even, but I could not be certain. I convinced myself not to google the formula, and set to thinking.

My mind immediately wandered to the idea of being relatively prime (or coprime). Two numbers, although they may not be prime, are coprime if they share no common factor. I knew that new triples could be “built” simply by multiplying each entry (a,b, and c) by some constant ‘m’ because:

if a^2 +b^2 = c^2
m(a^2 +b^2) = m(c^2)

I played around with common triples and new ones created by multiplying by certain constants, but I new that this was fruitless. Each triple I could think of had an even entry, therefore any multiple of that entry would also be divisible by 2. My time spent on this hypothesis was driven mainly by the intrigue that such a relationship must exist.

The existence of a formula paralyzed me. I new it was out there and that Euclid had done all the work for me. I could not bring myself to embrace the simplicity of the problem; I was attempting to find the general formula instead of showing that one entry must be even. Not only did the formula kill my desire to derive it, it closed down all other pathways to an answer.

Then, this morning, I had a spark of childlike thinking. It was so elementary that I smiled when it came to mind. (Elementary when compared to the derivation of Euclid’s Formula). I tell my students that mathematicians are lazy; I also make it quite clear that this doesn’t mean they do no work. It means that they always try to find the quickest and most succinct path from hypothesis to proof. Ideas should not be disposed as trivial because they are simple. Some of the greatest theorems of all time come from modest assumptions. With this spirit in mind, I sat down and answered the question.

My new angle was this:

Let’s create a generic triple from 3 odds, and show that it cannot be true

I had reduced the question to a series of trials–an extremely short series of trials. I then set-up a situation where 3 odd numbers satisfied the Pythagorean relationship.

a^2 + b^2 = c^2
a = 2x+1, b = 2y+1, c=2z+1 | x,y,z are integers
(2x+1)^2 + (2y+1)^2 = (2z+1)^2
4x^2 + 4x +1 + 4y^2 + 4y + 1 = 4z^2 +4z + 1
collect terms and factor out a 2…
2[2x^2 + 2x + 2y^2 + 2y + 1] = 2[2z^2 + 2z] + 1

Now because all of x,y, and z are integers, the expression “[2x^2 + 2x + 2y^2 + 2y + 1]” is also an integer. So the Left-hand side (LHS) is an even number. (The entire thing is divisible by 2). The Right-hand side (RHS) is an odd number because it is of the form 2n + 1, with ‘n’ an integer.

If we clean it up a bit by letting ‘m’ and ‘n’ represent general integers:

2m = 2n + 1

This relationship is not solvable over the set of the integers. It essentially states that if a Pythagorean triple didn’t contain an even entry, an Even number would have to equal an Odd number. This can’t happen. The official reason is because the Odds and Evens form a bijection. Not important…

What other problems can be posed from this fact?

Can we have a Triple with 2 Evens and 1 Odd?
Can the Even be a,b, or c? Does it matter?
Can a triple be all Evens? Can I find one?
What organizations of Evens and Odds satisfy the equation:
a^2 + b^2 + c^2 = d^2 ??

I didn’t need Euclid’s formula to engage in meaningful mathematics. In fact, the knowledge of a formula acted as a roadblock. The feeling of enlightenment is one that I want my students to experience when they do mathematics. It is these “aha” moments that fuel my drive to create meaningful lessons. They invigorate me even when a students make a connection that I have already made, but I assure you this is not always the case. In this situation, life without Euclid provided me the opportunity to make numerous connections.


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