proof of Diophantus' theorem on Pythagorean triples

The proof is quite easy and it comes down to formulae that you might recognise from calculus for the sine, cosine and tangent of a double angle.

Start with the unit circle v²+w²=1 centred at the origin and with radius one. We can give a closed formula for a point (v,w) on the circle. A typical point is
(v,w)=(1-t²) / (1+t²), 2t / (1+t²)) (*).

The proof of this assertion is a consequence of those trigonometric identities I mentioned but a more geometric point is given as follows (you will want to draw a diagram to understand this). We project the circle onto the w-axis. Consider the line that goes through the point (-1,0) and hits the w-axis at the point (0,t). Where does this line meet the circle? Well it meets it exactly at the point (*) above.

It's clear that this point (v,w) is a pair of rational numbers if and only if t is rational. So to get back to the proof of the theorem, suppose that we have a primitive Pythagorean triple (a,b,c). Then (a/c,b/c) is a point on the unit circle. So by the above argument there corresponds a rational number t to this point. Write t=x/y in its lowest terms (so that x and y are coprime). The formula (*) nows gives us that a=x²-y², b=2xy and c=x²+y². This shows that every primitive triple has the required form. That triples of the form in the statement are primitive Pythagorean triples is clear, so we are done.

To repeat the assertion, it is that any primitive Pythagorean triple (a,b,c) will be equal to (x^2-y^2,2xy,x^2+y^2) for two natural numbers x and y. First, consider the fact that c must be either even or odd. For the moment, let c be even. Then either a and b can both be even, or both be odd. If a, b and c are all even, then they do not form a primitive triple (they're all divisible by two), so we disregard this case. So a = 2j+1, b = 2k+1 and c = 2m. a^2 + b^2 = c^2, so (2j+1)^2 + (2k+1)^2 = (2m)^2. This yields 4j^2 + 4j + 1 + 4k^2 + 4k + 1 = 4m^2. But this statement says that an integer of the form 4n+2 is equal to an integer of the form 4p, and that's impossible. So, it's apparent that no primitive Pythagorean triple has an even hypotenuse.

Since c must be odd, a and b must be of different parity. Since it doesn't matter, let a be odd and b be even. Rewriting the Pythagorean theorem, we get b^2 = c^2-a^2 = (c-a)(c+a). We already decided that b is even, so we divide it by 2 and get (b/2)^2 = (c-a)/2 * (c+a)/2.

Obviously enough, c = (c+a)/2 + (c-a)/2, and a = (c+a)/2 - (c-a)/2. So if (c+a)/2 and (c-a)/2 had any common divisors, these divisors would also divide both c and a. But (a,b,c) is a primitive triple, so gcd((c+a)/2,(c-a)/2)) must be 1. Now, I don't know about you, but where I come from, when the product of two numbers is a square and the numbers don't share any divisors, we conclude pretty swiftly that the two numbers have to be squares themselves. And (b/2)^2 is a square, dammit! So we write that (c+a)/2 = x^2, and (c-a)/2 = y^2. We can now write c = x^2 + y^2, and a = x^2 - y^2. Finally, we have (b/2)^2 = x^2*y^2, so b/2 = x*y, and b = 2xy. QEFD