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Pythagorean triples and Hilbert's Theorem 90
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\\
Noam D. Elkies
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The classical parametrization of Pythagorean triples is well known:

{\bf Theorem.}
{\em
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 \hbox{$z^2-y^2 = x^2$}
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.  

{\em Proof}\/: We may assume that $z\neq 0$,
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 $a \in (\Q(i))^*$
such that $w = a/\overline a$.  For any nonzero $r\in\Z$
we have $ar/\overline{ar} = a/\overline a$.  There exists
nonzero $r\in\Z$ such that $ar \in \Z[i]$, say
$ar = m+in$ with $m,n\in\Z$, 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.

{\em 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 $z\neq 0$.  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 $a\in(\Q(r))^*$,
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 $w = (x+ry)/z \in \Q(c)$ 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 $z\0\neq 0$,
then we know $w\0 \in \Q(c)$ such that $w\0 \overline w\0 = C$,
namely $w\0 = (x\0 + r y\0)/z\0$.
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.

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