The area of a right triangle whose sides are integers in not a square implies ===>

Hello,

I have been trying to show that the fact that S1="The area of a right triangle whose sides are integers in not a square" implies that S2="There is no triangle whose sides are rational and whose area is 1".

So far I have that if we assume S1, and also consider an arbitrary triangle whose sides are rational and whose area is 1, then we have that:

$\displaystyle \frac{1}{2}\frac{x}{y}\cdot h=1\Rightarrow h=\frac{2y}{x},$ which shows that the height of one of these such triangles must be rational. Then, if we let $\displaystyle h=a/b$, we have that:

$\displaystyle \frac{1}{2}\frac{x}{y}\frac{a}{b}=1\Rightarrow \frac{1}{2}ax=by$

I have also considered the fact that our triangle is two right triangles put together, and that our triangle may be enclosed within a rectangle, but so far nothing has panned out.

The book treats this implication as trivial, saying that "if such a triangle existed, we would be able to obtain, by multiplying all three sides by a suitable integer, a triangle whose sides are integers and whose area is a square..." And I can see this, but I cannot see that this is a __right__ triangle, as is needed to contradict S1.