Show that no Pythagorean triangle can have its area equal its hypotenuse One problem we were given in our number theory course was to show that no Pythagorean triangle can have its area equal its hypotenuse.
Here is my attempt:
Let the sides of the triangle be given by: $$x=u^2-v^2, \quad y=2uv,\quad z=u^2+v^2,$$ where $u>v\geq1$, and $u,v$ are of opposite parity.
Suppose that the area of this triangle was equal to its hypotenuse. Then, $$\frac{1}{2}xy=z \implies uv(u^2-v^2)=u^2+v^2 \implies \left(\frac{u}{v}\right)^2=\frac{uv+1}{uv-1}.$$
My reasoning is that this can never be the case, since the fraction on the right hand side is of the form $(n+2)/{n}$ for $n \in \mathbb{N}$, and hence never square.
Is this reasonable?
 A: For fun we give a proof that uses no number-theoretic machinery.
Let the sides of our Pythagorean triangle be $x,y,z$, with $z$ the hypotenuse. If the area is equal to the hypotenuse, then $\frac{1}{2}xy=z$, or equivalently $2xy=4z$. Then from $x^2+y^2=z^2$ we obtain $$(x+y)^2=x^2+y^2+2xy=z^2+4z.$$ 
It follows that $z^2+4z$ is a perfect square. We show this is impossible.
The first perfect square after $z^2$ is $(z+1)^2$, which cannot be equal to $z^2+4z$. The next one, $(z+2)^2$, is greater than $z^2+4z$. So if $z$ is a positive integer, then $z^2+4z$ cannot be a perfect square. 
Remark: We did not use the representation theorem for Pythagorean triples. Your idea to use the theorem is good. There is a little problem. Not all Pythagorean triples are of the shape you described. For example, $(9,12,15)$ is not. To represent all triples, we need to multiply the triples you described by an arbitrary positive integer constant $k$. We can take $u$ and $v$ relatively prime and of opposite parity.
A: Let $b = 3$; $c = \frac {9}{\sqrt 5}$; $a = \frac {2c} b = \frac {6} {\sqrt 5}$
Then $a^2 + b^2 =  \frac {36}5 + 9 = \frac {81}{5} = c^2$
So it is possible for the area of a triangle to equal the hypotenuse.
