Re Michael Reid's proof, I had earlier sent Lunnon email explaining a proof idea which I believe coincides with Reid's proof idea (I did not have a proof, but I did have this underlying idea), I now excerpt from our correspondence: WDS-->Lunnon: If your triangle vertices are (0,0) (x1, y1) and (x2, y2) all integers, then regard the points as COMPLEX NUMBERS and take G = gcd( x1+y1*i, x2+y2*i ) as a Gaussian integer. We then may divide all 3 vertices by G. This does both a scaling and a rotation, and it scales it down by the most it possibly can (no?). Lunnon-->WDS: Aren't you assuming the result that you're trying to prove here? WDS-->Lunnon: --no, I begin with an integer embedding that I already gave formulas for every coordinate. The only problem is it just might not be primitive. To make it primitive we divide by a real GCD; to find the smallest lattice version we divide by a complex GCD. Lunnon then said, and I agreed, that we ought to check Reid's proof by computer before agreeing with it. About that, here is some computer output I made: Program for messing with Heronian Triangles. Warren D Smith Nov 2011 n=2 m=3 k=1 a=15 b=20 c=25 G=5 semip=30 area=150 xC=9 yC=12 xA=25 GIgcd=3+4*i norm=5 n=2 m=8 k=1 a=40 b=130 c=150 G=10 semip=160 area=2400 xC=24 yC=32 xA=150 GIgcd=6+8*i norm=10 n=2 m=13 k=1 a=65 b=340 c=375 G=5 semip=390 area=9750 xC=39 yC=52 xA=375 GIgcd=3+4*i norm=5 n=2 m=18 k=1 a=90 b=650 c=700 G=10 semip=720 area=25200 xC=54 yC=72 xA=700 GIgcd=6+8*i norm=10 n=3 m=7 k=1 a=70 b=150 c=200 G=10 semip=210 area=4200 xC=56 yC=42 xA=200 GIgcd=8+6*i norm=10 n=3 m=10 k=2 a=130 b=312 c=338 G=26 semip=390 area=20280 xC=50 yC=120 xA=338 GIgcd=10+24*i norm=26 n=3 m=12 k=1 a=120 b=435 c=525 G=15 semip=540 area=18900 xC=96 yC=72 xA=525 GIgcd=12+9*i norm=15 n=4 m=11 k=2 a=220 b=500 c=600 G=20 semip=660 area=52800 xC=132 yC=176 xA=600 GIgcd=12+16*i norm=20 n=4 m=11 k=3 a=275 b=520 c=525 G=5 semip=660 area=69300 xC=77 yC=264 xA=525 GIgcd=4+3*i norm=5 n=4 m=13 k=1 a=221 b=680 c=867 G=17 semip=884 area=45084 xC=195 yC=104 xA=867 GIgcd=15+8*i norm=17 n=5 m=8 k=1 a=208 b=325 c=507 G=13 semip=520 area=20280 xC=192 yC=80 xA=507 GIgcd=12+5*i norm=13 n=6 m=9 k=2 a=360 b=510 c=750 G=30 semip=810 area=81000 xC=288 yC=216 xA=750 GIgcd=24+18*i norm=30 n=7 m=8 k=1 a=400 b=455 c=825 G=5 semip=840 area=46200 xC=384 yC=112 xA=825 GIgcd=3+4*i norm=5 n=7 m=8 k=4 a=520 b=560 c=600 G=40 semip=840 area=134400 xC=264 yC=448 xA=600 GIgcd=32+24*i norm=40 To explain the lines of the table, the Buchholz parameters are n,m,k yielding the (not primitive) triangle with sides a,b,c. All Buchholz triangles with semiperim<1000 should be listed that do not have a trivially integer-embeddable primitivization. G is the integer GCD of (a,b,c). The coordinates of the naive integer embedding of the vertices A,B,C which I had derived in a previous post are A=(xA,0), B=(0,0), and C=(xC,yC). FInally, GIgcd is the Gaussian-integer GCD of (xA, xC+yC*i) and norm is its absolute value (modulus). You will notice an amazing coincidence of the G and norm columns and the amazing fact norm is always an integer. Lunnon now wonders: "What next? On to tetrahedra?" Well, maybe... I point out that in 4 dimensions, the "Hurwitz integers" are the quaternions A+B*i+C*j+D*k such that either (A,B,C,D) or (A+1/2, B+1/2, C+1/2, D+1/2) all are integers. http://en.wikipedia.org/wiki/Hurwitz_quaternion These enjoy a Euclidean-like GCD algorithm, albeit be careful since quaternion multiplication is noncommutative, Any 3D rotation+scaling can be expressed using quaternions as follows: Let (x,y,z), the 3-vector being rotated, be regarded as X=0+x*i+y*j+z*k and perform X --> Q*X*Qconjugate. [Also any 4D rotation+scaling can be expressed using quaternions as follows: X --> Q*X*R.] -- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step)