Why can’t there exist two Pythagorean triples with the same hypotenuse, but one has a leg that is twice as long as a leg of the other? Or why is there no integer solution to
$a^2 + b^2 = c^2$
$4a^2 + d^2 = c^2$
$a$ must be the even leg.
The existence would imply that $c - 2a$, $c - a$, $c + a$, $c + 2a$ are all squares.
So I think I have found a proof that no such triples exist:
Suppose such a solution exists, thus as noted in the question $c-2a,\ c-a,\ c+a,\ c+2a$ are all squares. Rescaling, we will have rational squares $r^2,\ s^2, 1, t^2$ where $r^2 = 1-3a$, $s^2 = 1-2a$ and $t^2 = 1+a$. Thus $(6t^2,6rst)$ is a rational point on the elliptic curve $y^2 = x^3-17x^2+72x$ which has rank $0$ (I checked LMFDB for this). We can now check each of the torsion points to see that none of them leads to a set of $r,s,t$ with an $a\neq 0$. (The torsion subgroup is generated by $(8,0)$ and $(6,6)$ according to LMFDB.) Of course there are ways to prove that the rank is $0$ and the torsion subgroup is as claimed without relying on computers but I decided to keep this short.