Can a regular $n$-simplex have vertices in $\mathbb Z^n$ for $n > 1$?

Trivially, a regular $$0$$-simplex (point) and $$1$$-simplex (line segment) can have integer vertices in $$0$$ and $$1$$ dimensional Euclidean space respectively. On the other hand, a regular $$2$$-simplex (equilateral triangle) cannot have vertices in $$\mathbb Z^2$$.

Is it true more generally that an $$n$$-simplex cannot have vertices in $$\mathbb Z^n$$ for $$n > 1$$? Is there an easy way to see this?

Edit:

Eugen J. Ionascu provides an example of a regular tetrahedron with vertices in $$\mathbb Z^3$$: $$(0,0,4),(7,0,3),(3,5,0),(4,5,7)$$

Do such regular $$n$$-simplices exist for all $$n \neq 2$$?

• What about $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$? – Michael Biro Feb 20 at 1:29
• See one of the answers in mathoverflow.net/questions/38724/…. The statement is that it's possible iff n+1 is the sum of one, two, four or eight odd squares! – Michael Biro Feb 20 at 1:36
• It looks like the question is answered here on MathOverflow. – Peter Kagey Feb 20 at 1:45