Lunnon: The parameterisation given by Buchholz (spelling!) --- which he ascribes to Carmichael, incidentally --- does not guarantee a primitive triangle with GCD(edge-lengths) = 1, only some triangle similar to it. --so let me understand your question. What you want is a proof that every PRIMITIVIZED heronian triangle is integer-posable? I.e. you conjecture any Bucholz triangle AFTER dividing by GCD(a,b,c) is integer-posable? Call that "revised Lunnon conjecture." To repeat the formulas so far Bucholz: a := (n*n + k*k)*m; b := (m*m + k*k)*n; c := (m + n)*(m*n - k*k); semiperim=s = (m + n)*m*n, area=d = k*m*n*(m + n)*(m*n - k*k) = k*m*n*c; constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m*n*n/(m + 2*n) ) vertex coordinates proposed by WDS: A=(+n*(m-k)*(m+k), 0) B=(-m*(n-k)*(n+k), 0) C=(0, 2*k*m*n) a^2 = |B-C|^2 = |B|^2 + |C|^2, b^2 = |A-C|^2 = |A|^2 + |C|^2, c^2 = |B-A|^2. Well clearly G = GCD(a,b,c) = sqrt(GCD(a^2, b^2, c^2, area)) divides a, and b, and c, and 2*s=2*(m+n)*m*n; and G^2 divides area. If G divides 2*k*m*n=|C|, then G^2 divides |A|^2 and |B|^2 and |C|^2 and a^2 and b^2 and c^2 and area forcing G to divide a, b, c, A, B, and C. So "revised Lunnon" <== "G must divide 2*k*m*n." But we know G divides GCD(2*s, c, a+b). Hence G divides (m+n)*GCD(2*m*n, m*n-k*k, m*n+k*k). Hence G divides (m+n)*GCD(m*n, 2*k*k). But Lunnon-constraint implies that GCD(m*n,2*k*k)=GCD(m*n,2*k). Hence G divides 2*k*m*n. Therefore, the revised Lunnon conjecture, is also proven. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)