Oct 4 Assignment 1 Solutions and Extension

To find positive integer solutions to Pythagorean triples, it is proposed that the ancient Babylonians used the reciprocal pairs method: Given s (some fraction greater than 1, which they wrote in sexagesimals) and 1/s we take their sum and difference, s+1/s and s-1/s. Next we take the difference of their squares: (s+1/s)^2 - (s-1/s)^2 which is always equal to 4 = 2^2. This was a great finding since the equation can be rearranged to (s-1/s)^2 + 2^2 = (s+1/s)^2. However this doesn't necessarily give postive integer solutions to x^2 + y^2 = z^2. To see how what could be done, since s is some fraction, write s as p/q where p>q>0. Then (s-1/s)^2 + 2^2 = (s+1/s)^2 becomes [(p^2-q^2)/pq]^2 + 2^2 = [(p^2+q^2)/pq]^2. Multiply both sides by (pq)^2 to get (p^2-q^2)^2 + (2pq)^2 = (p^2+q^2)^2. Given this formula now, positive integers p and q can be chosen such that x=p^2-q^2, y = 2pq, z = p^2-q^2. The problem is that this won't always generate a reduced Pythagorean triple. If the GCD of x,y,z is not 1, it'll have to be divided out from each of the three in order to get a reduced triple. Several observations/conditions were gleaned regarding the conditions for p and q to get a reduced triple, but no proof has been made on those claims. Modern solutions use more areas in known number theory to derive a proof on how to generate reduced Pythagorean triples. The conditions are that p and q are coprime, and exactly one of them is even and the other odd. This is a necessary and sufficient condition in order to have reduced triples. 

Reduced Pythagorean triples satisfy the Pythagorean equation x^2 + y^2 = z^2, where x,y denote the length of the two legs of a right triangle and z its hypotenuse length. The Pythagorean theorem, which applies to right triangles, is a special case of the cosine law, a^2+b^2-2abcos(C) = c^2, where the angle opposite to the hypotenuse, c, is 90°. Since cos(90°) = 0, it boils down to just a^2+b^2=c^2. The cosine law is useful for finding an angle measure when given all side lengths. It is also useful for finding a missing side when given the other sides and one angle measure. A great proof video on the derivation of the cosine law can be found on Khan Academy.