I finally caught up with Michael Reid's proof, and at least understand HOW his algorithm works --- even if I still have only the faintest notion WHY ... Essentially it computes a magic complex number via Gaussian integer GCD, representing a rotation about a chosen vertex at the origin, through which a chosen edge along the axis is rotated to the desired lattice pose. There are 6 ways to choose the vertex and edge, 4 Gaussian units to vary the GCD, and 2 conjugations of the rotation (yes or no), yielding in general 48 different possible results. If the triangle possesses one axial and one non-axial lattice pose, each can be translated, rotated and rotated trivially in 24 variations, yielding again 48 results. While I can't quite put my finger on exactly why, I think the proof must accommodate the following stronger result. Conjecture: (modulo trivial symmetries) right, axial non-right, non-axial Heronian triangles have respectively just one axial, one axial plus one non-axial, one non-axial pose. On 11/18/11, Michael Reid <reid@gauss.math.ucf.edu> wrote:

Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain.

Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' .

Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer.

Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED

Note that this gives a constructive way to find such an embedding.

Michael Reid