A right triangle with natural numbered sides and area $=abcd$

Question

Let $a,b,c$ and $d$ be natural numbers such that $a+b=c$ and $a+d=2c$ .

Prove that there is a right triangle with natural-numbered sides and area $=abcd$.

My try :

$x^{2}+y^{2}=z^{2}$ and $S=\frac{x+y}{2}$

But I don't have no ideas to continue.

Answer 1

We require a right angle triangle whose area is $abcd=ab(a+b)(a+2b)$ (using the constraints.)

The lengths of a right angle triangle can be paramerised by \begin{eqnarray*} x&=& \alpha^2- \beta^2 \\ y&=& 2 \alpha \beta \\ z&=& \alpha^2+ \beta^2. \\ \end{eqnarray*} This triangle will have area $ \alpha \beta (\alpha- \beta)(\alpha+ \beta)$. Now choose $ \alpha= \beta+\gamma$ and the area becomes $ \beta \gamma( \beta +\gamma)(2 \beta+\gamma)$. So the area of every right angle triangle has the aforementioned form!