= Solution
Let $P_0=(-4,6)$, the infinite-order point established in part (c). Every positive multiple $nP_0$ 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 $y\ne0$ and $x\notin\{0,5,-5\}$.
The stated algebraic identity gives positive rational sides
$$
\boxed{a=\frac{|x^2-25|}{|y|},\qquad b=\frac{|10x|}{|y|},\qquad c=\frac{x^2+25}{|y|}.}
$$
They satisfy $a^2+b^2=c^2$, and their area is
$$
\frac{ab}{2}=\frac{5|x(x^2-25)|}{y^2}=5
$$
by 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 \b[there are infinitely many distinct rational right triangles of area five]. For $P_0$ itself the sides are $(3/2,20/3,41/6)$.
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