The elliptic curve $C_1:Y^2=X^3+pX$ has rank $0$ where is $p$ is a prime equivalent to $7$ mod $16$. This is exercise 3.8b) in the Silverman-Tate book “Rational Points on Elliptic Curves”.. In the book by Silverman “The Arithmetic of Elliptic Curves” he proves it in Proposition 6.2. This is not (to my mind) an elementary proof. But the Silverman-Tate book uses only elementary methods, so there should be an elementary proof.
For folks who don't have the book, here is how the book suggests to solve problems of this type:
- There is an isogenous curve $C_1: Y^2=X^3-4pX$, with isogenies in each direction whose composition is multiplication by $2$.
- Solutions on $C$ or $C_1$ to equations of the form $N^2=b_1M^4+b_2e^4$ where $b_1b_2 =$ the $X$ coefficient give rise to points on the curve $C$.
- These points allow the rank to be computed. The book has the details. Since the rank is supposed to be $0$, we expect the equations for $M$ and $e$ non-zero to have no solutions. I haven't been able to prove this for some of the equations, for example the equation $N_2=4 M^4- p e^4$ It should be possible to prove no solutions in an elementary way, or to derive the rank of $C$ in some other elementary way, but I don't see it. Am I missing something? It is not even a starred problem.
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: