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?
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.