Show that either a or b must be odd and the other even-

Question

Consider Pythagorean triples (a, b, c) with a² + b² = c², and with a, b, c ∈ Z. Suppose moreover that a and b have no common factors.

(a) Show that either a or b must be odd, and the other even.

(b) Show in this case (assuming a is odd and b even) that there are integers m, n so that a = m² - n², b = 2mn, and c = m² + n². [Hint: Note that b² = (c - a)(c + a), and prove that (c - a)/2 and (c + a)/2 are relatively prime integers.]

(c) Conversely, show that whenever c is a sum of two-squares, then there exist integers a and b such that a² + b²= c².