TITLE: Can a parallelogram have whole-number lengths for all four sides and both diagonals?
QUESTION [20 upvotes]: Is it possible for a parallelogram to have whole-number lengths for all four sides and both diagonals?

One idea I had was to arrange four identical right triangles such that the right angles are adjacent. For example if we take four triangles that are all 3-4-5 right triangles, and arrange them so that their legs form a cross with two arms that are 3 units and two arms that are 4 units, their hypotenuses will form a parallelogram in which all sides are 5 units – that is, a rhombus.
A second idea was to take two pairs of isosceles triangles with all legs the same length, where one vertex angle is supplementary to another, and arrange them so that their vertex angles are adjacent. For example we could take two triangles with two sides of 6 units that form a 25° angle, and another two triangles with two sides of 6 units that form a 155° angle, and arrange them to form a quadrilateral with 12-unit diagonals – that is, a rectangle.
That's all I can imagine. My hunch is that a parallelogram can have whole-number lengths for all four sides and both diagonals only if it is either a rhombus or a rectangle.
Is that right? If so, can this limitation be explained elegantly?

REPLY [6 votes]: For a quadrilateral of sides $a,b,c,d$, diagonals $D_1,D_2$ and $m$ the distance between the midpoints of these diagonals it is known the formula
$$a^2+b^2+c^2+d^2=D_1^2+D_2^2+4m^2$$
For a parallelogram we have $a=c$ and $b=d$ and $m=0$ so one has the formula
$$2(a^2+b^2)=D_1^2+D_2^2$$
On the other hand  for the equation $X^2+Y^2=2Z^2$ the general solution with $(X,Y)=1$ is given by the identity
$$(r^2-s^2+2rs)^2+(r^2-s^2-2rs)^2=2((r^2-s^2)^2+(2rs)^2)$$ where $Z=r^2+s^2$ complete in the RHS a pythagorean triple.
Consequently making
$$\begin{cases}a=r^2-s^2\\b=2rs\\D_1=r^2-s^2+2rs\\D_2=r^2-s^2-2rs\end{cases}$$ we can get infinitely many examples of the required parallelograms.