Each of the two diagonals and each of the four sides of a rhombus has integer lengths, satisfying the condition: x^{2} + y^{2} = 4z^{2}, where x, y are the length of the diagonals and z being the length of each side.

Is the area of the rhombus always divisible by 24?

If so, prove it. Otherwise provide a counter example.

Let x=2a and y=2b

Than a,b,z form a Pythagorian triple .(PT).

A primitive PT has * inter alia* two features:

- Exactly one of
*a*, *b* is divisible by 3 - Exactly one of
*a*, *b* is divisible by 4

So ab is 12*pq.- p,q integers.

The word **exactly **might be redundant in non primitive PT.

S=.5*XY= .5*4*ab=,5*4*12pq=24pq

Ergo :

**the area of the "integer" rhombus**

** is ****always divisible by 24.**