I will use the notations from loc. cit., so let us consider the curves
$$
\begin{aligned}
C &= C_p\qquad & y^2 &= x^3 + px\ ,\\
\bar C &= \bar C_p\qquad & y^2 &= x^3 -4px\ ,
\end{aligned}
$$
with $p$ a prime congruent to $7$ modulo $16$.
We want to show that the rank $r$ of $C$ is zero, so that $\Gamma :=E(\Bbb Q)$ has only the torsion part of order two, $\Gamma =\{O,T\}$, $O$ being the neutral element, and $T=(0,0)$. Recall the morphism $\alpha$ from $\Gamma$ to $\Bbb Q^\times $ modulo squares, mapping a generic point $(x,y)$ to $x$ modulo squares, and $O\to 1$, $T\to p$ modulo squares. Its image is a subgroup of $\langle -1,p\rangle$ modulo squares. We know that $p$ is in the image. Is $-1$ also in the image? If yes, then there is some rational $x=-m^2\ne 0$ with a corresponding $y$ satisfying the equation of $C$, this is not possible in reals (at the real place), so we know $\alpha(\Gamma)=\langle p\rangle$ has order two.
There is a similar morphism $\bar\alpha:\bar\Gamma\to(\Bbb Q^\times$ modulo squares$)$, also mapping a point $(\bar x,\bar y)\in \bar\Gamma :=\bar C(\Bbb Q)$ to its first component, taken modulo squares. We also need the exact description of $\bar\alpha(\bar\Gamma)$, which is a subgroup of
$\langle -1,2,p\rangle$ modulo squares.
The last group has eight elements, so $\bar\alpha(\bar\Gamma)$ has order among $1,2,4,8$. The $1$ is excluded, since the class of $-4p=-p$ is in the image of $\bar\alpha$, $\bar T=(0,0)$ is mapped to this element.
Recall now that every non-torsion element $\bar P=(\bar x,\bar y)$, $\bar x\ne 0$, is of the shape
$$
\begin{aligned}
\bar P&=(\bar x,\bar y)=\left(\ \frac{b_1M^2}{e^2}\ ,\frac{b_2MN}{e^3}\ \right)\ ,\\
&\qquad\text{ and there are the relatively prime pairs of numbers:}\\
1&=(M,e)=(N,e)=(b_1,e)=(M,b_2)=(M,N)\ ,\\[3mm]
\bar\alpha(\bar P)&=\bar x=b_1\text{ modulo squares ,}\\
b_1b_2&=-4p\ ,\\
N^2 &= b_1M^4 + b_2e^4\ .
\end{aligned}
$$
Which $b_1$-values cannot be realized? We note first that $-1$ and $-2$ are not quadratic residues modulo $p$, using the information on $p$ modulo $16$. So the equations:
$$
\begin{aligned}
N^2 &= -M^4 + 4p\;e^4\ ,\\
N^2 &= -2M^4 + 2p\;e^4\ ,\\
N^2 &= 4p\;M^4 -e^4\ ,\\
N^2 &= 2p\;M^4 -2\;e^4\ ,
\end{aligned}
$$
have no solutions in integers $\ne 0$,
so $-1$, and $-2$, and their "opposites" (w.r.t. $\bar b=-4p$) $p\equiv 4p$, and $2p$ are not in the image of $\bar\alpha$.
(The first two not being in the image imply the other two are also not images, since $-4p$ is in the image.) So the image of $\bar\alpha$ has either two, or four elements, we need to exclude one more element among $2$, $2p$. Take the $2$, so we want to solve in integers $M,N,e$ with the above meanings:
$$
N^2 = 2M^4 -2p\;e^4\ .
$$
So $N$ is even. We consider the above relation modulo $16$. The LHS is using either $N$ of the shape $N=4k$, or of the shape $4k+2$, so $N^2$ is either $0$, or $4$ modulo $16$. The RHS has odd values for $M,e$ (because they are relatively prime to $N$). So $M$ is of the shape $4s\pm 1$, giving with the binomial formula $M^4\equiv (\pm1)^4 =1$ modulo $16$. Same game for $e$. Then $2M^4 -2p\; e^4=2-2\cdot 7=2-14=-12=4$ modulo $16$. So $N$ is of the form $N=2N_1=2(4k\pm1)$. From $2N_1^2=M^4-2p\; e^4$ we obtain modulo $16$ the information $2=2\cdot (\pm1)^2=1-7=-6$ modulo $16$, contradiction.
In other words, the equation fails, because it fails at the place $2$.
We know now the image of $\bar\alpha$. Finally, the rank $r$ of $\Gamma =C(\Bbb Q)$ is given by (see loc. cit. at the bottom of page $91$)
$$
2^r=\frac 14\cdot \#\alpha(\Gamma)\cdot \#\bar\alpha(\bar\Gamma)
=\frac 14\cdot2\cdot 2=1\ ,
$$
giving $r=0$.
It may be interesting to compare the above information with the information delivered by an engine, sage here, for a special choice of $p$, say $p=23$, below we have more or less the same game - and i asked sage to give me the information with maximal verbosity:
sage: E = EllipticCurve(QQ, [23, 0])
sage: E
Elliptic Curve defined by y^2 = x^3 + 23*x over Rational Field
sage: E.rank()
0
sage: E.simon_two_descent(verbose=3)
ellQ_ellrank([0, 0, 0, 23, 0], []);
starting ellQ_ellrank
Elliptic curve: Y^2 = x^3 + 23*x
E[2] = [[0], [0, 0]]
Elliptic curve: Y^2 = x^3 + 23*x
Algorithm of 2-descent via isogenies
computing the 2-torsion
E[2] = [2, [2], [[0, 0]]]
Search for trivial points on the curve
Y^2 = x^3 + 23*x
trivial solution: lead(pol) is a square
trivial solution: pol(0) is a square
trivial solution: roots of pol[0]
point found by ratpoint = [0, 0]
trivial points on E(Q) = [[0, 0], [1, 1, 0], [0, 0], [0, 0]]
#K(b,2)gen = 1
K(b,2)gen = [23]~
d1 = 23
quartic = y^2 = 23*x^4 + 1
trivial solution: pol(0) is a square
point on the quartic
[0, 1]
points on E(Q) = [[0, 0]]
[E(Q):phi'(E'(Q))] = 2
#S^(phi')(E'/Q) = 2
#III(E'/Q)[phi'] = 1
#K(a^2-4b,2)gen = 3
K(a^2-4b,2)gen = [-1, 2, 23]~
computing the 2-torsion
E[2] = [2, [2], [[0, 0]]]
Search for trivial points on the curve
Y^2 = x^3 - 92*x
trivial solution: lead(pol) is a square
trivial solution: pol(0) is a square
trivial solution: roots of pol[0]
point found by ratpoint = [0, 0]
trivial points on E'(Q) = [[0, 0], [1, 1, 0], [0, 0], [0, 0]]
d1 = -1
quartic = y^2 = -x^4 + 92
not ELS at 2
d1 = 2
quartic = y^2 = 2x^4 - 46
not ELS at 2
d1 = -2
quartic = y^2 = -2x^4 + 46
not ELS at 2
d1 = 23
quartic = y^2 = 23x^4 - 4
not ELS at 2
d1 = -23
quartic = y^2 = -23x^4 + 4
trivial solution: pol(0) is a square
point on the quartic
[0, 2]
points on E'(Q) = [[0, 0]]
points on E(Q) = [[0, 0]]
[E'(Q):phi(E(Q))] = 2
#S^(phi)(E/Q) = 2
#III(E/Q)[phi] = 1
#III(E/Q)[2] = 1
#E(Q)[2] = 2
#E(Q)/2E(Q) = 2
rank = 0
Reduction of the generators [[0, 0], [0, 0]]
computing the 2-torsion
E[2] = [2, [2], [[0, 0]]]
points = [[0, 0]]
end of ell2descent_viaisog
end of ellQ_ellrank
v = [0, 1, [[0, 0]]]
Rank determined successfully, saturating...
(0, 1, [])
sage:
