How to prove that it is possible to make rhombuses with any number of interior points? I was given some square dot paper which can be found on this link: http://lrt.ednet.ns.ca/PD/BLM/pdf_files/dot_paper/sq_dot_1cm.pdf  and was told to draw a few rhombuses with the vertices on the dots but no other dots on the edges.
Here is an example:
This is not a rhombus it just shows how the shape must be lay out. 

                   A = 6

The number of interior dots would be represented by A. I was then told that it is possible to draw rhombuses with any number of interior points. All I had to do was prove this.
I started by trying to draw these shapes and came up with some of the below:

A = 2

A = 4

A = 9  
However, I am still unsure of how you can prove something like this since there may be numerous ways of drawing these rhombuses.
I would appreciate support.
Thank you :) 
 A: If $l$ is the desired number of internal lattice points, for odd $l$ take the rhombus defined by the points 
$(0,0), ({l+1\over 2}, {l-1\over 2}), ({l-1\over 2}, {l+1 \over 2}), (l, l)$ 
and for even $l$ the rhombus defined by the points
$(0,0), ({l\over 2}+1, -{l\over 2}), (-{l\over 2}, {l\over 2}+1), (1, 1)$.
Explanation:
A rhombus may be defined by two vectors $(a,b)$ and $(c,d)$ of equal length by placing the vertices at $(0,0), (a,b), (c,d),$ and $(a+c, b+d)$.  Any parallelogram defined in this manner is known by basic linear algebra to have area $|ad-bc|$.  By relabeling we may assume $ad-bc > 0$, so the area is simply $ad-bc$.
Now assume $a,b,c,d \in \mathbb{Z}$.  The segment from $(0,0)$ to $(a,b)$ will have no lattice points on it except the endpoints iff $\gcd (a,b) = 1$, since if $(m,n)$ is another lattice point on the same line we must have $m/n = b/a$ (the slope of the line measured two ways).  But this means $b/a$ wasn't in lowest terms.  i.e. $\gcd (a,b) \ne 1$.  The same argument works for the segment from $(0,0)$ to $(c,d)$, and the other two segments will then not contain any lattice points besides vertices since they're just translations of the first two.  Our conditions so far on $a,b,c,d$ are
(a) $a^2+b^2 = c^2+d^2$,
(b) $\gcd(a,b) = \gcd(c,d) = 1$.
By Pick's theorem the number of internal lattice points in a lattice polygon is $l = A-b/2+1$, where $A$ is the area and $b$ is the number of lattice points on the boundary.  For us, $A = ad-bc$ and $b=4$, so we find
(c) $l+1 = ad-bc$
The coordinates for even $l$ come from guessing $a=d, b=c$ and solving
$l+1 = a^2-b^2 = (a-b)(a+b)$ 
by setting $a-b=1, a+b=l+1$.
This doesn't work for odd $l$, but we can do this instead.  Multiply (c) by $2i$ and add it to (a), then rearrange to get
$(a+id)^2-(c+ib)^2 = 2i(l+1)$
divide by $i$ and factor to get
(d) $[(a+c)+(b+d)i][(-b+d)+(-a+c)i] = 2(l+1)$
Thus $a,b,c,d$ lead to a factorization of $2(l+1)$ in $\mathbb{Z}[i]$.  The coordinates at the top come from solving (d) for $a,b,c,d$ in the factorization
$2(l+1) = [(l+1)+(l+1)i][1-i]$
Finally, in both cases the condition (b) on gcds is quickly checked, since in all cases the coordinates of $(a,b)$ or $(c,d)$ differ by 1 in absolute value, and so must be relatively prime.
A: The square dot paper you describe can be thought of as something called the integer lattice, that is, the set of points $(m,n)$ in the plane where $m,n$ are integers. In order to make this identification, pick an arbitrary point on the square dot paper and call it $(0,0)$. If $m,n$ are positive integers, then $(m,n)$ is the point which is $m$ points to the right and $n$ points up from $(0,0)$. If $m$ is negative, move to the left instead of the right. If $n$ is negative move down instead of up.
To rephrase what we wish to prove with this terminology, we could say that for any number of points $s$, there is a rhombus $R$ with vertices on the integer lattice whose boundary (except for the vertices) contains no points of the integer lattice, and $R$ contains $s$ points of the integer lattice in its interior.
Your first example generalizes naturally in the following way.
Let $n$ be a positive integer and consider the rhombus $R$ whose vertices consist of the points $(1,0)$, $(-1,0)$, $(0,n)$ and $(0,-n)$. The rhombus $R$ contains no points of the integer lattice on its boundary except its vertices since its sides have $x$-values between $0$ and $1$. Moreover, $R$ contains the points $(0,k)$ for $-n < k < n$ within its interior, of which there are $2n-1$. This proves that you can find a rhombus containing a desired odd number of points of the integer lattice in its interior. If you are having trouble seeing this, try drawing a rhombus like this on your dot paper.
To find a rhombus which contains a desired even number of points of the integer lattice in its interior, try using a diagonally placed rhombus like yours in the first example (note: that one has 2 points in the interior). Have a shot at working out the rest of this argument for yourself; it is similar to the one I have above.
A: I'm a student who was assigned this question as well, so I figured that reading an explanation written by another peer would be more beneficial for you. I have used pictures and examples to aid you in understanding the concept, rather than using nonsensical mathematics. I hope this helps!
We are easily able to construct a rhombus with any number of interior points if we consider these 2 patterns.

Let R equal "rhombus" and I equal "Interior points," (referring to the diagram) If we construct rhombuses at this particular angle, we can acquire any amount of EVEN NUMBERED (2,4,6,8 etc) interior points, we just have to expand the rhombus by 1 point each time on both sides. I have highlighted the points of each rhombus relative to each other to make it clearer for you.
After some observation, you can see a clear pattern emerge, but the problem is, what about odd numbers? I have another pattern similar to the even numbered one, but the rhombuses are just at a different angles.

As you can see, if we just tilt the rhombus at a different angle, we are able to acquire ODD NUMBERED (1,3,5,7 etc) Interior points. Same as before, we reach higher odd numbers by expanding the rhombus by a point on each side. I have highlighted each rhombuses' points relative to each other to make it clearer for you.
Now, let's combine these 2 patterns to prove the question.
0 Interior points = R1 (first picture)--------
1 Interior point  = R1 (second picture)--------
2 Interior points = R2 (first picture)--------
3 Interior points = R2 (second picture)--------
4 Interior points = R3 (first picture)--------
5 Interior points = R3 (second picture)--------
Although we have run out of examples, can you clearly see a pattern? This pattern can continue to any number of interior points you desire.
Therefore, we have proved the theory of "constructing a rhombus with any amount of interior points." You can directly use my answer, or you can reword it yourself, I don't mind.
I hope that my answer assisted you with your geometry problem! I aimed to construct my response as clearly as possible, so hopefully you understood my explanation.
A: The proof is straight-forward. Lets use the property that rhombus is a quadrilateral whose diagonals are perpendicular bisectors. We'll accommodate all the interior points we want along a line. 


*

*Odd number of interior points


Say we want $p_o$ interior points; consider $p_o$ consecutive points along a line (notice the square paper has two symmetries - straight and diagonal, so we can choose either a straight or a diagonal line. Lets choose straight line). Now take one point each on either side of these $p_o$ points; they shall serve as two of the vertices of our desired rhombus. Consistent with our definition, choose the point which bisects the distance between these $p_o+2$ points; from the point of bisection, draw a perpendicular line. Immediate points on each side that this line intersects are the other two vertices of our rhombus. Join the four vertices and we have our desired rhombus. 


*Even number of interior points


We follow the same procedure, but since the line perpendicularly bisecting the distance between two $p_e+2$ points will never intersect an other point on the square paper, we make use of its diagonal symmetry. So, we choose our $p_e$ points along a diagonal instead and follow the same procedure.
So, the proof follows from Rhombus's diagonals being perpendicular bisectors.
A: For many values of $n$, there are a lot more ways to draw a rhombus around $n$  points than you need. There are really only two ways you need to think about drawing a rhombus according to the given rules; these two ways of drawing the rhombus will let you enclose any non-negative number of lattice points
(points with integer coordinates) that you want.
The first kind of rhombus looks like this:

Choose an integer $m$. If you suppose the points on your paper represent all
the points with integer coordinates in a Cartesian plane,
put the leftmost vertex of the rhombus at the point with coordinates $(0,0)$,
then put the other three points at $(m,1)$, $(2m, 0)$,
and $(m, -1)$ respectively.
the figure above shows the result when $m = 3$.
You can show that


*

*two adjacent edges of this quadrilateral are of equal length;

*each pair of opposite edges are equal and parallel;

*each edge lies between the lines $y=0$ and $y=1$ (or $y=-1$ and $y=0$);

*each edge has lattice points at its endpoints and nowhere else;

*the only lattice points that can be in the interior of the rhombus are the points on the line $y=0$;

*lattice points $(x,y)$ inside the rhombus satisfy $0 < x < 2m$;

*there are exactly $2m - 1$ lattice points inside the rhombus.


The other kind of rhombus looks like this:

This time, after you choose an integer $m$, you
put the leftmost vertex of the rhombus at the point with coordinates $(0,0)$,
then put the other three points at $(m,m+1)$, $(2m+1, 2m+1)$,
and $(m+1, m)$ respectively.
the figure above shows the result when $m = 2$.
You can show that


*

*two adjacent edges of this quadrilateral are of equal length;

*each pair of opposite edges are equal and parallel;

*each edge lies between the lines $y=x$ and $y=x+1$ (or $y=x-1$ and $y=x$);

*each edge has lattice points at its endpoints and nowhere else;

*the only lattice points that can be in the interior of the rhombus are the points on the line $y=x$;

*lattice points $(x,y)$ inside the rhombus satisfy $0 < x < 2m+1$;

*there are exactly $2m$ lattice points inside the rhombus.


Then you can prove that for any non-negative integer of the form $2m-1$,
where $m$ is an integer, you can enclose $2m - 1$ points in a rhombus;
and for any non-negative integer of the form $2m$,
you can enclose $2m$ points in a rhombus.
The only remaining thing to show is that every non-negative integer $n$
can be written either as $n=2m-1$ or as $n=2m$, where $m$ is an integer.

By the way, if you look at either of the two earlier answers carefully, 
you may see that they tell you to use essentially 
the same series of rhombuses (possibly shifted up, down, and/or sideways
and possibly rotated $90$ degrees, none of which changes the results
in any way that matters)
to enclose any odd number of lattice points 
and any even number of lattice points.
Consider that when assigning credit; each of those answers may merit
the award, or at least part of it.
