Introduction to Pythagorean Triples
The Pythagorean Theorem tells us that a right triangle with legs a and b and hypotenuse c will always have the relationship a2+b2 = c2. (Do you know how to prove that?) When all three sides are whole numbers, we have a Pythagorean triple. The most famous of these is 32+42 = 52, often referred to in this context as (3,4,5). The 3-4-5 triangle was used in Egypt to help make perpendicular sides for their magnificent buildings. A loop of rope with 12 equally spaced knots (3+4+5 = 12) was pulled taut at knots 0, 3 and 7 to make a precise right angle.
If the three sides don't all have a factor in common, then they make a primitive Pythagorean triple (PPT). 62+82 = 102 is not a PPT because all three sides have a factor of 2.
Starting to Explore
When I'm making up problems for students, I often want another Pythagorean triple, so I have one more sitting in my brain, 52+122 = 132. Are there others? Are there infinitely many PPT's? How would we find more?
One approach to exploring these involves thinking about parity. (Parity refers to whether a number is odd or even.) I see that both of the PPT's above, (3,4,5) and (5,12,13), are odd+even = odd. Can we have odd+odd = even? What about even+even = even? If we have odd+even = odd, does the odd leg have to be the shorter one?
Do other questions occur to you?
Recommendation: stop reading and start playing as soon as you have a thought about how you might proceed.
I knew there were more PPT's, but couldn't remember any others. I wanted to find a few more, so I could see any obvious patterns. So I made a list of the first 25 perfect squares and looked for pairs that would add to equal another perfect square, or subtract to equal another one. I found (8,15,17) and (7,24,25). Well, that's one question answered: the odd leg does not have to be the shortest side. I see that all 4 hypotenuses are odd. So I want to address the question of whether odd+odd = even is possible.
Question1: Is odd+odd = even possible?
Here's how I start: Suppose we have two odd legs. We can let a = 2n+1 and b=2m+1. Then a2+b2 = ... Can we come to a contradiction? [See the hint at the end for a bit more direction.]
Question2: If a is an odd number, can I find a PPT for it?
I notice that all 3 triples in which the odd leg is the short one include consecutive numbers for the other leg and the hypotenuse. Hmm. If a is the short leg, then b+c = a2 in all 3 of those cases. Is that important? What if I write a2+b2 = c2 as c2-b2 = a2? Oh! A square minus a square can factor. So I'd have (c-b)(c+b) = a2. What does that get me? [I found a way to get a PPT for every odd number. Can you?]
Question3: Must the even side be a multiple of 4?
I wanted more triples, in case it would help me see more patterns. So I set up a spreadsheet with column a holding 1 through 100, row 1 holding 1 through 254 (first time I've ever used all available rows!). Column b and row 2 had the squares of these numbers. The rest of the spreadsheet showed a 0 if the square root of the sum of these squares was not a whole number, and otherwise showed the number. [Here's the formula in cell c3: =IF(INT(SQRT($B3+C$2))=SQRT($B3+C$2),SQRT($B3+C$2),0)]
For most multiples of 4, I found a PPT. And I didn't find any for the other even numbers. So I wanted to know whether the even side had to be a multiple of 4. If the even side is a, then b and c are odd, and c2-b2, with b=2m+1 and c=2n+1, can be explored.
Questions 4 and 5: Multiples of 3, 4 and 5
This reminded me that I had read (whose blog was that on?) that in every PPT, 3 will be a factor of one side, 4 will be a factor of one side, and 5 will be a factor of one side. (As in (5,12,13), one side may contain more than one of these factors.) I realized I'd already proved it for 4. I started trying to prove it for 3.
I mentioned parity earlier. Even numbers can be expressed as a=2m, and odd numbers can be expressed as b=2n+1. Similarly, if we want to think about whether side c will be a multiple of 3, we can look at three cases: c=3m, c=3m+1, or c=3m+2. Using this, I started with the question of whether c would be a multiple of 3. (It wasn't in any of the triples I'd found.) If it is, then neither a nor b can be. (Why?) Once I solved that problem, I wanted to prove that one of the legs would be a multiple of 3. Suppose b is not a multiple of 3 and consider c2-b2. There will be 4 cases, each of b and c can either be 3x+1 or 3x+2. What does this make a?
I tried to think about 5 in this way but got nowhere. I'm writing this blog post in hopes that explaining my thinking will help me get further on some of my dead ends.
I have a few other questions I haven't answered:
- Given a multiple of 4, how can I come up with a PPT?
- Can the same number show up in 3 different PPT's?
Hint: c2 is a multiple of 4. (Why?) What about a2+b2?