Pythagorean Triple - Generating A Triple - Proof of Euclid's Formula

Proof of Euclid's Formula

That satisfaction of Euclid's formula by a, b, c is sufficient for the triangle to be Pythagorean is apparent from the fact that for integers m and n, the a, b, and c given by the formula are all integers, and from the fact that

A simple proof of the necessity that a, b, c be expressed by Euclid's formula for any primitive Pythagorean triple is as follows. All such triples can be written as (a, b, c) where a2 + b2 = c2 and a, b, c are pairwise coprime, and where b and c have opposite parities (one is even and one is odd). (If c had the same parity as both legs, then if all were even the parameters would not be coprime, and if all were odd then a2 + b2 = c2 would equate an even to an odd.) From we obtain and hence . Then . Since is rational, we set it equal to in lowest terms. We also observe that equals the reciprocal of and hence equals the reciprocal of, and thus equals . Then solving

for and gives

Since and are fully reduced by assumption, the numerators can be equated and the denominators can be equated if and only if the right side of each equation is fully reduced; given the previous specification that is fully reduced, implying that m and n are coprime, the right sides are fully reduced if and only if m and n have opposite parity (one is even and one is odd) so that the numerators are not divisible by 2. (And m and n must have opposite parity: if both were odd then dividing through by 2 would give the ratio of two odd numbers; equating this ratio to, which is a ratio of two numbers with opposite parities, would give conflicting parities when the equation is cross-multiplied.) So equating numerators and equating denominators, we have Euclid's formula with m and n coprime and of opposite parities.

A longer but more commonplace proof is given in Maor (2007) and Sierpinski (2003).