Can a rectangle in the plane contain exactly 51 lattice points?

Question

The flag of the United States has a blue field with 50 stars. These stars are roughly on a 2D integer lattice (on point). If the US were to get another state (say Puerto Rico or something) then they would presumably update the flag to include 51 stars. I was wondering if it is possible to have a field of stars still roughly on the integer lattice and contain 51 exactly without some obvious hole. Still with a rectangular border.

Any ideas? Clearly you could do a $(51 \times 1)$ box but that's rather uninteresting. I was hoping to find something more exotic. I worry that this may be off topic, but it seems mathematical enough in nature. I thought I'd ask here. I can't figure this out, but I have no idea whether this is easy or hard.

Answer 1

Currently the rows go $6+5+6+5+6+5+6+5 +6= 50$, a total of $9$ rows.

You can make the rows $9+8+9+8+9+8=51$, which has $6$ rows.

The way you can find solutions is by setting up the Diophantine equation $n[x+(x+1)]=51$, or $n(2x+1) = 51$. You can solve this by replacing $51$ with $3 \cdot 17$ or $1 \cdot 51$ and going through the posibilities. You can change it up to any number of stars, and to $x+(x+a)$ for any $a$ if you wish.

Picture from Wikipedia:

Answer 2

Here's a stranger way of doing it...

If the integers $x$ and $y$ are coprime, then there is a rectangle containing exactly the lattice points $(0,0)$ and $(x,y)$. (Have the rectangle's major axis pass through both of these points and go just a little bit beyond them. Then make the rectangle narrow enough that it doesn't include any other points.).

We can extend this rectangle along its length to get a rectangle which includes exactly the points $(0,0), (x,y), (2x,2y), \ldots, (50x, 50y)$.