Proof for Pythagorean Solutions

If a² + b² = c² and a, b, and c are relatively prime, then there exists integers p and q such that a = p² – q², b = 2pq, and c = p² + q² where p and q are relatively prime.

001	Assume a2 + b2 = c2, where a, b, and c are relatively prime.
002	Since a, b, and c are relatively prime, either only one of a, b, and c is even or none of a, b, and c are even.
003	Assume none of a, b, and c are even:
004	Then a, b, and c are all odd
005	And c is even since c2 = a2 + b2 and an odd plus and odd is an even.
006	But this contradicts that c is even.
007	So it is not the case that none of a, b, and c are even.
008	Since either only one of a, b, and c is even or none of a, b, and c are even, and it is no the case that none of a, b, and c are even, that means one of a, b, and c is even.
009	Assume c is even:
010	Then c = 2C
011	And a and b must be odd such that a = 2A + 1 and b = 2B + 1
012	So c2 is divisible by 4 because c2 = (2C)2 = 4C2
013	And a2 + b2 is not divisible by 4 because a2 + b2 = (2A + 1)2 + (2B + 1)2 = 4A2 + 4A + 1 + 4B2 + 4B + 1 = 4(A2 + A + B2 + B) + 2
014	But since a2 + b2 = c2, c2 is both divisible by 4 and not divisible by 4, which is a contradiction.
015	So c is not even.
016	So either a or b are even, and since the formula is symmetrical, we will choose b as even, and a and c as odd. Since b is even, let b = 2u, where u is an integer
017	Since a and c are odd, c + a is an even (since an odd plus an odd is an even), so c + a = 2w, where w is an integer.
018	Also, c – a is an even (since an odd minus an odd is an even), so c – a = 2v, where v is an integer.
019	Now u2 = vw, because u2 = ¼(4u2) = ¼(2u)2 = ¼b2 = ¼(c2 – a2) = ¼(c – a)(c + a) = ¼(2v)(2w) = vw.
020	And c = w + v, because c = ½(2c) = ½(c + c) = ½(c + c + a – a) = ½(c + a + c – a) = ½((c + a) + (c – a)) = ½(2w + 2v) = w + v.
021	And a = w – v, because a = ½(2a) = ½(a + a) = ½(a + a + c – c) = ½(c + a – c + a) = ½((c + a) – (c – a)) = ½(2w – 2v) = w – v.
022	Assume v and w are not relatively prime:
023	Then v = fV and w = fW
024	And a is divisible by f, since a = w – v = fW – fV = f(W – V)
025	And c is divisible by f, since c = w + v = fW + fV = f(W + V)
026	But this contradicts that a and c are relatively prime.
027	So v and w are relatively prime.
028	Since u2 = vw, and v and w are relatively prime, v and w must be squares as well, so w = p2 and v = q2, where v and w are integers.
029	So c = p2 + q2, because c = w + v = p2 + q2
030	And a = p2 – q2, because a = w – v = p2 – q2
031	And b = 2pq, because b = √(b2) = √(c2 – a2) = √((p2 + q2)2 – (p2 – q2)2) = √((p4 + 2p2q2 + q4) – (p4 – 2p2q2 + q4)) = √4p2q2 = 2pq.
032	Assume p and q are not relatively prime:
033	Then p = fP and q = fQ
034	And a is divisible by f2 since a = p2 – q2 = (fP)2 – (fQ)2 = f2P2 – f2Q2 = f2(P2 – Q2)
035	And c is divisible by f2 since c = p2 + q2 = (fP)2 + (fQ)2 = f2P2 + f2Q2 = f2(P2 + Q2)
036	But this contradicts that a and c are relatively prime.
037	So p and q are relatively prime.
038	Therefore, if a2 + b2 = c2 and a, b, and c are relatively prime, then there exists integers p and q such that a = p2 – q2, b = 2pq, and c = p2 + q2 where p and q are relatively prime.

Q.E.D.