Consider the 4-space simplex PQRST having edge-lengths QP = 1, RP = sqrt(2), RQ = sqrt(2), SP = sqrt(3), SR = sqrt(3), SQ = sqrt(3), TP = 2, TQ = sqrt(2), TR = 1, TS = 1. Let lattice vertices be labelled even / odd according to the sum of their Cartesian coordinate components. The triangle PQR has sides 1,sqrt(2),sqrt(2): if it is embedded in the lattice with P is even then Q is odd, so R is odd, so P is odd. By contradiction, it must be impossible to embed PQRST in the lattice |Z^4. However, PQRST does have the rational embedding in |Q^4 P = [0,0,0,0], Q = [0,0,0,1], R = [1,0,1,0], S = [1,1,0,0], T = [5,1,1,0]/3. Why am I telling you this? The proofs that a Heronian simplex is lattice embeddable in 2 and 3 dimensions require only that the simplex be rationally embedded, and have edge lengths squaring to integers. This example shows that any proof of Heronian embeddability in |Z^4 would have to somehow involve extra properties besides these two. [That's always assuming, of course, that somebody can manage to find any to embed: Sascha Kurz has recently calculated that no Heronian pentatope can exist with diameter < 600,000.] Incidentally, I have prepared a proof of the construction of lattice embeddings in 2- and 3-space via complex-number and quaternion GCD. I'll try to upload this somewhere public shortly --- in the meantime, anyone interested is welcome to request a copy (don't all rush at once!). Fred Lunnon