= Solution
{c}
We use <two-isogeny descent>, stating the arithmetic results involved. For
$$
E:y^2=x(x^2+ax+b),\qquad
E':Y^2=X(X^2-2aX+a^2-4b),
$$
define the <two-torsion square-class homomorphism> $\alpha_E:E(\mathbb Q)\to\mathbb Q^\times/\mathbb Q^{\times2}$ by $\alpha_E(o)=1$, $\alpha_E((0,0))=b$, and $\alpha_E((x,y))=x$ otherwise. Its kernel is the image of the dual <two-isogeny formula>; the analogous map $\alpha_{E'}$ has kernel the image of the forward <two-isogeny formula>. The <two-isogeny rank formula> is
$$
2^r=\frac{|\alpha_E(E(\mathbb Q))|\,
|\alpha_{E'}(E'(\mathbb Q))|}{4}.
$$
The images have signed squarefree representatives supported on <primes> dividing $b$ and $a^2-4b$, respectively. This <prime-support bound in two-isogeny descent> follows by <valuation> comparison: if a <prime> does not divide $b$, the <valuations> of $x$ at that <prime> must be even, whether positive or negative. A candidate squarefree class $d$ is represented precisely when its <quartic covering in a two-isogeny descent>
$$
N^2=dU^4+aU^2V^2+(b/d)V^4
$$
has the corresponding rational solution, apart from the separately included identity and two-torsion cases; clearing denominators permits <coprime> <integers> $U,V$.
Here $a=5,b=-5$, so
$$
E':Y^2=X(X^2-10X+45).
$$
On $E$ the possible classes are $1,-1,5,-5$. They all occur: the identity gives one, $T=(0,0)$ gives $-5$, and $(-1,3)$ gives $-1$. Since the image is a <group>, their product also gives five. Thus its image has order four.
On $E'$, the quadratic factor $(X-5)^2+20$ is strictly positive for real $X$. Every nonzero rational $X$-coordinate is therefore positive. The <prime-support bound in two-isogeny descent> restricts its image to $1,3,5,15$. The identity gives one and $(0,0)$ gives $45$, that is class five.
Class three is impossible. Its <quartic covering in a two-isogeny descent> would require
$$
N^2=3U^4-10U^2V^2+15V^4,\qquad\gcd(U,V)=1.
$$
Modulo three this is $N^2\equiv2U^2V^2$, so three divides $N$ and at least one of $U,V$. They cannot both be divisible. If $3\mid U$ and $3\nmid V$, the right side has <valuation> exactly one, from $15V^4$. If $3\mid V$ and $3\nmid U$, it again has <valuation> exactly one, from $3U^4$. Either case contradicts its being a square. Class fifteen is then impossible too, since multiplying it by the known image class five would give class three.
Hence the second image is exactly $\{1,5\}$, of order two. The rank formula gives
$$
2^r=\frac{4\cdot2}{4}=2,\qquad\boxed{r=1.}
$$
This proves the <rank-one elliptic curve with coefficients five and minus five> calculation, including the local obstruction that removes the spurious extra descent classes.
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