The two factors and of a primitive Pythagorean triple each equal the square of a Gaussian integer. This can be proved using the property that every Gaussian integer can be factored uniquely into Gaussian primes up to units. (This unique factorization follows from the fact that, roughly speaking, a version of the Euclidean algorithm can be defined on them.) The proof has three steps. First, if "a" and "b" share no prime factors in the integers, then they also share no prime factors in the Gaussian integers. (Assume "a" = "gu" and "b" = "gv" with Gaussian integers "g", "u" and "v" and "g" not a unit. Then "u" and "v" lie on the same line through the origin. All Gaussian integers on such a line are integer multiples of some Gaussian integer "h". But then the integer "gh" ≠ ±1 divides both "a" and "b".) Second, it follows that "z" and "z*" likewise share no prime factors in the Gaussian integers. For if they did, then their common divisor δ would also divide "z" + "z*" = 2"a" and "z" − "z*" = 2"ib". Since "a" and "b" are coprime, that implies that δ divides 2 = (1 + i)(1 − i) = i(1 − i)2. From the formula "c"2 = "zz*", that in turn would imply that "c" is even, contrary to the hypothesis of a primitive Pythagorean triple. Third, since "c"2 is a square, every Gaussian prime in its factorization is doubled, i.e., appears an even number of times. Since "z" and "z*" share no prime factors, this doubling is also true for them. Hence, "z" and "z*" are squares.