TITLE: Lattice n-gons with ordered side lengths 1,2,3,...,n
QUESTION [13 upvotes]: Consider the octagon in the Cartesian plane with vertices at (0,0), (1,0), (1,2), (4,2), (4,6), (7,2), (7,8), and (0,8). 
Are there other (infinitely many) polygons, such as this, lying entirely in the first quadrant of the Cartesian plane with all its vertices at lattice points and whose sides, in anti-clock wise order, are precisely of lengths 1, 2, 3, ... n starting from the side departing at the origin?

REPLY [10 votes]: There are indeed other such polygons.
-- For example there is one for $n = 11$, as follows
(the origin is in the lower left corner):

Also there is one for $n = 15$:

Further there are $21$ such polygons for $n = 16$.
One of them is the following:

These polygons have been found with this GAP function.
There is one such polygon for $n = 8$, $n = 11$ and $n = 15$, each,
there are $21$ such polygons for $n = 16$, and $225$ for $n = 19$.
For all other $n < 20$ there are no such polygons.
The complete list for $n \leq 19$ in GAP-readable form
can be found here.
Added on May 3, 2016: A zip file with *.png images of all
$249$ polygons for $n \leq 19$ can be found here (545KB).
Added on May 4, 2016 (following a request in the comments):
When leaving away the condition that the polygon is entirely in the
first quadrant, we get $3$, $5$, $6$, $584$ and $882$ distinct polygons
for $n = 8$, $n = 11$, $n = 12$, $n = 15$ and $n = 16$, respectively.
A GAP-readable coordinate list can be found here, and
a zip file with *.png images of all these $1480$ polygons
can be found here (2.8MB).