= Solution
For $E:y^2=x^3-17x$, the <two-isogeny formula> gives $E':Y^2=X^3+68X$. By the <prime-support bound in two-isogeny descent>, the image of the <two-torsion square-class homomorphism> on $E$ lies in $\{1,-1,17,-17\}$. The identity gives $1$, $T=(0,0)$ gives $-17$, and $P=(-1,4)$ gives $-1$, since $(-1)^3-17(-1)=16$. The image is a <subgroup> of the <square-class group>, so their product $17$ occurs as well; explicitly $P+T=(17,68)$. Hence
$$
\alpha(E(\mathbb Q))=\{1,-1,17,-17\}.
$$
On $E'$, a nonzero affine point has $X>0$, because $Y^2=X(X^2+68)$. The only candidate <square classes> are therefore $1,2,17,34$. The identity gives $1$, $T'=(0,0)$ gives $[68]=17$, and $(2,12)$ gives $2$, since $2^3+68\cdot2=144$. Their product $34$ occurs; adding $T'$ gives $(34,-204)$. Thus
$$
\alpha'(E'(\mathbb Q))=\{1,2,17,34\}.
$$
Both images have order four. The <two-isogeny rank formula> proves
$$
\boxed{2^r=\frac{4\cdot4}{4}=4,\qquad \operatorname{rank}E(\mathbb Q)=2.}
$$
This is the <rank-two elliptic curve with cubic x cubed minus seventeen x>; all candidate classes are represented, so no unresolved <local-to-global gap in isogeny descent> remains in this computation.
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