= Solution
By part b, write
$$
E_5(\mathbb Q)=\mathbb ZG\oplus E_5[2].
$$
The <canonical height of an elliptic curve> vanishes on torsion and satisfies $\widehat h(mG+T)=m^2\widehat h(G)$. If $\widehat h(P)=\widehat h(Q)\ne0$, their nonzero integer coefficients therefore have equal squares, so
$$
Q=\pm P+T
$$
for some $T\in E_5[2]$.
Negation leaves the $x$-coordinate unchanged. Addition by the four 2-torsion points changes $x$ among
$$
x,qquad-\frac1x,qquad\frac{x+1}{x-1},
\qquad-\frac{x-1}{x+1}.
$$
Substitution in the three side formulas for $\Delta_P$, using $5y^2=x^3-x$, only changes signs and permutes the three values. Hence $\Delta_Q$ and $\Delta_P$ are the same right triangle up to reordering their sides.
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