= Solution
Complete the square-class map at the identity by $q(O)=1$; at the finite point $T=(0,0)$ its value is the class of $b$. For another <rational point> write
$$
x=r/s^2,\qquad y=t/s^3,\qquad\gcd(r,s)=1.
$$
These denominator powers follow because a negative valuation of $x$ forces the leading cubic term to dominate, so that valuation is even and the valuation of $y$ is three halves of it. The equation gives
$$
t^2=r(r^2+ars^2+bs^4).
$$
If a prime $\ell\nmid b$ divides $r$, it does not divide the second factor, since $\ell\nmid s$. Thus $v_\ell(r)$ is even. All denominator valuations are even too. Every image class therefore has a representative which is a signed squarefree divisor of $b$, proving that $\operatorname{im}q$ is finite.
Use the standard 2-isogeny with <kernel> $\{O,T\}$ and <dual isogeny> satisfying $\widehat\phi\phi=[2]$, $\phi\widehat\phi=[2]$. Away from the exceptional points one may take
$$
\phi(x,y)=\left(x+a+\frac bx,\ y\left(1-\frac b{x^2}\right)\right).
$$
Applying the analogous isogeny to $E'$ gives a curve with coefficients $4a,16b$; scaling its coordinates by $1/4,1/8$ gives the dual map back to $E$. The analogous descent map $q'$ on $D$ has <kernel> $\phi(C)$ and finite image. The sequence
$$
D/\phi(C)\ \xrightarrow{\widehat\phi}\ C/2C\longrightarrow C/\widehat\phi(D)\longrightarrow0
$$
is exact at the last two terms. Its outside groups are finite, identified with the two finite descent images, so \b[$C/2C$ is finite].
For $E_p:y^2=x(x^2-p^2)$, the rational <2-torsion> is $O,(0,0),(p,0),(-p,0)$, of order four. The first descent image is exactly $\{1,-1,p,-p\}$: these classes occur on the <2-torsion>, and the divisor-support bound permits no others. The isogenous curve is $E'_p:y^2=x(x^2+4p^2)$. Its real points have $x\ge0$, so the second image lies in the four positive classes $\{1,2,p,2p\}$.
Here the <kernel> $\{O,(0,0)\}$ of the dual is contained in $\phi(C)$, since $\phi(\pm p,0)=(0,0)$. Thus the displayed map $D/\phi(C)\to C/2C$ is injective, and exactness gives
$$
|C/2C|=|\operatorname{im}q|\,|\operatorname{im}q'|=4|\operatorname{im}q'|.
$$
The allowed rank formula makes the left side $2^{R+2}$, so
$$
\boxed{2^R=|\operatorname{im}q'|\le4,\qquad R\le2.}
$$
Now consider the quartic relation. Normalize a nontrivial <integer> solution so that $p$ does not divide both $x,y$, by dividing them by $p$ and $z$ by $p^2$ whenever necessary. If $p\nmid x$, reduction modulo $p$ gives $x^4=2z^2$, with $z\ne0$ modulo $p$, making $2$ a <quadratic residue>. If $p\mid x$, then $p\nmid y$ and $p\mid z$; write $x=pX$, $z=pZ$, divide by $p^2$, and reduce to get $y^4=2Z^2$ modulo $p$. Again $2$ is a square. Therefore
$$
\boxed{\left(\frac2p\right)=1,\qquad p\equiv\pm1\pmod8.}
$$
If the class $2$ occurred in $\operatorname{im}q'$, write the corresponding coordinate as $x'=2u^2/v^2$. The curve equation then gives
$$
u^4+p^2v^4=2z^2,\qquad z=\frac{y'v^3}{4u}.
$$
Here $u,v$ are nonzero coprime integers and $z$ is rational. Since $2z^2$ is an <integer>, its reduced denominator must be one, so $z$ is an <integer> too. For $p\equiv3,5\pmod8$ this contradicts the quartic obstruction. Hence class $2$ is absent. A <subgroup> of $\{1,2,p,2p\}$ omitting $2$ has order at most two, so $2^R\le2$ and $R\le1$.
For $p=5$, direct substitution verifies the nonintegral point
$$
\boxed{(x,y)=\left(\frac{25}{4},\frac{75}{8}\right).}
$$
A rational torsion point would have integral coordinates by the preceding theorem, so this point has infinite order. Thus $R\ge1$, and together with the upper bound,
$$
\boxed{\operatorname{rank}E_5(\mathbb Q)=1.}
$$
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