Originally Posted by

**earboth** Hello,

to 2): (I don' know how to "whiten" the Latex formulae, therefore I post my reply completely visible)

A right triangle with hypotenuse c and legs a and b has integer sides if

a = uČ - vČ, u > v and u, v in IN

b = 2uv

c = uČ + vČ

(for instance u = 2, v = 1 will give the famous 3, 4, 5 triangle)

The area of a right triangle is

$\displaystyle A=\frac{a \cdot b}{2}$

Plug in the above mentioned terms for a and b:

$\displaystyle A=\frac{(u^2+v^2) \cdot 2uv}{2}=(u^2-v^2) \cdot uv$

Now you have to prove that

- $\displaystyle u^2-v^2=(u+v)(u-v)$ is only a square if u=v or u=1 and v=0

- $\displaystyle u \cdot v$ is only a square if u = v

- $\displaystyle u^2-v^2=uv$. Solve this equation for u and you'll get:

$\displaystyle u=\frac{v \pm v \cdot \sqrt{5}}{2}$ that means u is not in IN

Thus the area can't be a square.

EB