 Solution

Let $P_0=(-4,6)$, the infinite-order point established in part (c). Every positive multiple $n{P}_0$ is rational and is neither $O$ nor 2-torsion: either possibility would force some nonzero multiple of $P_0$ to vanish. Write its coordinates as $(x,y)$, so $y\ne 0$ and $x\notin \{0,5,-5\}$.

The stated algebraic identity gives positive rational sidesThey satisfy $a^2+b^2=c^2$, and their area isby the curve equation. This is the infinitely many rational right triangles from a nontorsion point construction for the congruent number elliptic curve.

The infinitely many points cannot produce only finitely many triangles. For a fixed ratio $r=c/b$, the positive number $X=|x|$ satisfies $X^2-10rX+25=0$, so there are at most two choices for $|x|$, then at most two signs of $x$ and two signs of $y$. Thus the map to ordered triangles is finite-to-one. Permuting the two legs introduces only another finite ambiguity. Consequently there are infinitely many distinct rational right triangles of area five. For $P_0$ itself the sides are $(3/2,20/3,41/6)$.