 Solution

If $\sqrt{2}+\sqrt{3}$ were rational, squaring would make $\sqrt{6}$ rational, a contradiction. Alsoso it is irrational but algebraic, satisfying $x^2+2x-1=0$.

For lattice vertices $u,v,w$, the triangle area isa half-integer. Triangulating a convex lattice polygon from one vertex makes its area rational.

A regular octagon of side length $s$ has area $2(1+\sqrt{2})s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.