A bit of Pythagoras

Given Pythagorean triples that satisfy the Diophantine equation:

a² + b² = c²

where a, b and c share no common factors, one of a and b must be odd, the other must be even, and c is always odd.
First, note that squares of even numbers are always divisible by 4. [Lemma 1]

(2n)² = 4n²

And squares of odd numbers are of the form (4n + 1). [Lemma 2]

(2n + 1)² = 4n² + 4n + 1
= 4n(n+1) + 1

If a and b are both even, it is trivial to see that a, b and c share the factor 2.

c² = (2n)² + (2k)²
= 4n² + 4k²
= 4(n² + k²)

By Lemma 1, c is even. Therefore the solution reduces to the more primitive form:

n² + k² = (c/2)²

Consider a and b both odd:

c² = (2n + 1)² + (2k + 1)²
= 4n² + 4n + 1 + 4k² + 4k + 1
= 4n² + 4n + 4k² + 4k + 2
= 2(2(n² + n + k² + k) + 1)

This (by Lemmas 1 & 2) is not a perfect square, so this cannot be a solution. In fact, in this case c must be irrational:

Let p = n² + n + k² + k
c = √2√(2p + 1)

2p + 1 cannot have 2 as a factor. And since √2 is irrational, so is c. Therefore one of a and b must be odd and the other even.

c² = (2n + 1)² + (2k)²
= 4n² + 4n + 1 + 4k²
= 4n² + 4n + 4k² + 1
= 4(n² + n + k²) + 1

And c is odd (Lemma 2). QED.

One Response to “A bit of Pythagoras”

  1. Dawn says:

    You’re giving me Abstract Algebra flashbacks 🙂

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