Another rule about Pythagorean triples.

Last January, I wrote several posts about Pythagorean Triples, a set of three whole numbers a, b and c that follows the rule of the Pythagorean Theorem a² + b² = c². Here are some examples.

3² +4² = 5²
8² + 15² = 17²
24² + 7² = 25²

If we think about these numbers as sides of a triangle, the area will be ½ab. In the cases from above, the areas as 6, 60 and 84, respectively.

There is a way to generate all such whole number triples, using two positive integers j and k, with j>k.

a = j² - k²
b = 2jk
c = j² + k²

Our first rule is that we should stipulate that j does not equal k, because in the that case a = 0. Re-writing a and b in the area formula into multiples of j and k, we get this.

Area = ½ab = ½(j² - k²)2jk = (j² - k²)jk

Since j and k are whole numbers, the area must be a whole number. But more than that the number must be divisible by 6. The proof is split into two parts, first that the area is divisible by 2 and secondly that the area is divisible by 3.

Proof of the area is divisible by 2. If j or k is even, then (j² - k²)jk is even, since if you have a product of whole numbers and one is even, the product is even. The only other option is that both j and k is odd, and if that is the case then j² - k² = odd - odd = even.

Proof of the area is divisible by 3. If j or k is a multiple of 3, then (j² - k²)jk is a multiple of 3. The only other option is that neither j and k is a mulitple of 3. For any number n that isn't a multiple of 3,  n² will be of the form 3p + 1, which is to say one more than some multiple of 3. If that is the case then j² - k² = (3p + 1) - (3q + 1) = 3(p- q), which is a multiple of 3 as well.