Question

In class we proved that if (x, y, z) is a primitive Pythagorean triple, then (switching x and y if necessary) it must be that (x, y, z) = (m2 − n 2 , 2mn, m2 + n 2 ) for some positive integers m and n satisfying m > n, gcd(m, n) = 1, and either m or n is even. In this question you will prove that the converse is true: if m and n are integers satisfying these conditions, then (m2 − n 2 , 2mn, m2 + n 2 ) is a primitive Pythagorean triple.

(a) Show algebraically that the triple (m2 −n 2 , 2mn, m2 +n 2 ) satisfies the equation x 2 +y 2 = z 2 , so it is a Pythagorean triple.

(b) Prove that if m and n are relatively prime but not both odd, then the numbers {m2 − n 2 , 2mn, m2 + n 2} are mutually relatively prime. [Hint: Argue by contradiction, assuming that p is a common prime factor. Consider the cases p = 2 and p 6= 2 separately.]

Answer 1