= Solution
If $\sqrt2+\sqrt3$ were rational, squaring would make $\sqrt6$ rational, a contradiction. Also
$$
\tan(\pi/8)=\frac{1}{\sqrt2+1}=\sqrt2-1,
$$
so it is irrational but algebraic, satisfying $x^2+2x-1=0$.
For lattice vertices $u,v,w$, the triangle area is
$$
\frac12|\det(v-u,w-u)|,
$$
a half-integer. Triangulating a convex <lattice polygon> from one vertex makes its area rational.
A regular octagon of side length $s$ has area $2(1+\sqrt2)s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.
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