Finding integer coordinates for a pentagon, hexagon, heptagon, octagon, and nonagon, etc. Wondering what the formula is for finding integer coordinates for an arbitrary "regular" polygon. By regular I mean symmetrical polygons like pentagon, hexagon, etc.
In particular, I would like to know what integer coordinates are for a:


*

*Pentagon

*Hexagon

*Heptagon

*Octagon

*Nonagon


For instance, the Wikipedia pentagon SVG has these coordinates:
<polygon points="294,3 585.246118,214.602691 474,556.983037 114,556.983037 2.753882,214.602691" fill="white" stroke="black" stroke-width="4"/>

They are complex and unintuitive floating point values. I would instead like to figure out how to find purely integer coordinates, so for a pentagon it would be something like:
<polygon points="10,10 20,20 30,20 20,30 20,10" fill="white" stroke="black" stroke-width="4"/>

That doesn't make a pentagon, but it has all integer coordinates.
In order to accomplish this, it probably requires specifying a viewport or aspect ratio. So maybe it is wxh as 1000x1200, or 901x817 or something. I don't know how to figure this out.
To summarize, wondering what the equation is for figuring out integer coordinates for a symmetrical polygon, specifically those 5 above.
 A: Those don't exist, at least not in the 2-dimensional Euclidean plane. I'm 100% sure for all of the given ones except the octagon (EDIT: see at the end why the octagon is also impossible). The reason is that for 3 points $P,Q,R$ to be consecutive vertices of such a regular $n$-gon, the $\angle PQR$ must have size $180°-360°/n$. 
The tangens of the inclination $\alpha$ any straight line with equation $y=mx+n$ in a right-angled $(x,y)$-coordinate system has with the $x$-axis has is given by 
$$\tan \alpha = m$$
If that line goes trough 2 points with integer coordinates (say $P=(p_x, p_y)$ and $Q=(q_x,q_y)$), we get
$$\tan \alpha_{PG} = m_{PQ}=\frac{q_y-p_y}{q_x-p_x},$$
which implies that $m_{PQ}$ is a rational number.
The same goes for the line through $Q$ and $R$ and it's inclination $\alpha_{QR}$, its tangens is also a rational number. 
Now, the $\angle PQR$ we considered before is now just the difference of those two inclinations:
$$\angle PQR = \alpha_{PQ}-\alpha_{QR}$$.
The addition theorem for the tangens-function now says:
$$\tan \angle PQR = \frac{\tan \alpha_{PQ} - \tan \alpha_{QR}}{1+\tan \alpha_{PQ}\tan \alpha_{QR}}$$
The main point is now that if $P,Q,R$ have interger coordinates, this expression for $\tan \angle PQR$ is a rational number, as all the terms are rational and they are combined with elementatary operations that are closed for rationals numbers.
But $\tan 180°-360°/n$ is not a rational number for $n=5,6,7,9$. It is rational for $n=8$, as $180°-360°/8 = 135°$, so this proof does not work here.
I'm not sure what the answer is for $n=8$, could be either way.
ADDITION: 
It's impossible for the octagon as well. If $P,Q,R$ are the consecutive vertices of the octagon (so $Q$ is the middle one of the three), then $\angle QPR = 22.5°$. But $\tan 22.5°=\sqrt{2}-1$ is irrational, so the same argument as above applies, just for a different angle.
