Pythagorean triples and Hilbert’s Theorem 90

Noam D. Elkies

The classical parametrization of Pythagorean triples is well known:

Theorem. Integers x, y, z satisfy the Diophantine equation x^{2} + y^{2} = z^{2} if and only (x, y, z) is proportional to (m^{2} −n^{2}, 2mn, m^{2} +n^{2}) for some integers m, n.

This is usually proved either arithmetically, by rearranging the equation as z^{2} − y^{2} = x^{2} and factoring both sides, or geometrically, by the usual procedure for parametrizing a plane conic. It seems not to have been recognized that this parametrization of Pythagorean triples is equivalent to a special case of Hilbert’s Theorem 90.

Proof : We may assume that z ≠ 0, because if z = 0 then x = y = 0 and we may take (m, n) = (0, 0). Let w, then, be the complex number (x + iy)/z in the quadratic extension Q(i) of Q. The norm of w equals (x^{2} + y^{2})/z^{2} = 1. Thus by Hilbert there exists a ∈ (Q(i))^{∗} such that w = a/\overline{a}. For any nonzero r ∈ Z we have ar/\overline{ar} = a/\overline{a}. There exists nonzero r ∈ Z such that ar ∈ Z[i], say ar = m + in with m, n ∈ Z, not both zero. We then calculate

\frac{x + iy}{z} = w = ar/\overline{ar} = \frac{m + in}{m − in} = \frac{(m + in)^{2}}{(m + in)(m − in)} = \frac{(m^{2} − n^{2}) + i(2mn)}{m^{2} + n^{2}},

which is to say that (x, y, z) is proportional to (m^{2} − n^{2}, 2mn, m^{2} + n^{2}), as claimed.

Generalizations. In the same way we can parametrize the solutions of the Diophantine equation x^{2} + Axy + By^{2} = z^{2} for any constants A, B such that the discriminant A^{2} − 4B is not a square. Again we may assume that z ≠ 0. Instead of Q(i) we work in the quadratic extension Q(r) of Q, where r is a solution of r^{2} − Ar + B = 0. Then x^{2} + Axy + By^{2} is the norm of x + ry, so x^{2} + Axy + By^{2} = z^{2} if and only if w := (x + ry)/z has norm 1. Thus again w = a/\overline{a} for some a ∈ (Q(r))^{∗}, where \overline{a} denotes the algebraic conjugate of a. Hence

\frac{x + iy}{z} = w = \frac{m + rn}{m + \overline{r}n} = \frac{m + rn}{m + (A − r)n} = \frac{(m + rn)^{2}}{m^{2} + Amn + Bn^{2}}

for some integers m, n, not both zero. We conclude that (x, y, z) is proportional to (m^{2} − Bn^{2}, 2mn + An^{2}, m^{2} + Amn + Bn^{2}).

More generally yet, we may consider x^{2} + Axy + By^{2} = Cz^{2} for some nonzero constant C, with A, B as before. We then want w = (x + ry)/z ∈ Q(c) with norm w\overline{w} = C. In general such w need not exist. But if we know one solution (x_{0}, y_{0}, z_{0}) of x^{2} + Axy + By^{2} = Cz^{2} with z_{0} ≠ 0, then we know w_{0} ∈ Q(c) such that w_{0}\overline{w}_{0} = C, namely w_{0} = (x_{0} + ry_{0})/z_{0}. Then w\overline{w} = C if and only if w/w_{0} has norm 1, so w = w_{0}a/\overline{a} and we can proceed as before.