Mordell equation with prime-power constant

Question

(This question is somewhat related, but is different from this earlier question.)

I am interested in a specific case of the Mordell equation: $$E : y^2 = x^3 + k$$ where $k=q^t$, for some prime $q \equiv 1 \pmod 4$ and $t \equiv 1 \pmod 4$. Also, I will restrict $x$ to be even. It follows that $y$ is odd.

Question #1: Following the lead in the hyperlinked MSE question, does it follow that $E(\mathbb{Q})$ is isomorphic to $\mathbb{Z}/(t+1)\mathbb{Z}$?

I do not know how to answer Question #1. Any hints would be appreciated.

By the Nagell-Lutz theorem, we can narrow down the possibilities to $$y \in \{\pm 1, \pm 3, \pm q, \pm 3q, \pm q^2, \pm 3q^2, \ldots, \pm q^{t-1}, \pm 3q^{t-1}, \pm q^t, \pm 3q^t\}$$ since the discriminant of $E$ computes to $$D = -27q^{2t}$$ based from the Wikipedia hyperlink.

Question #2: How do I eliminate these possibilities? Any hints would be appreciated.

Answer 1

First, notice that an elliptic curve $E/K$ given by \begin{equation}\tag{$\star$} E : y^2 = x^3 + k \end{equation} is isomorphic to the curve $$E' : y^2 = x^3 + u^6 k$$ for any $u \in K^*$ by replacing $x$ by $u^2x$ and $y$ by $u^3y$. Hence by removing $6^{th}$ powers we are looking at

$$E : y^2 = x^3 + q^t$$

where $t = 1, 3, 5$ (in your case - I will deal with all $t = 0, ..., 5$ below).

Now, there is no reason to believe that when $q \equiv 1 \pmod{4}$ that $E/\mathbb{Q}$ will have rank $0$, and in particular when $q = 5$ and $t = 1$ the rank is $\geq 1$ (it is harder to show that equality holds) since $P = (-1, 2)$ has infinite order (this can be seen since e.g., $3P$ is some horrific fraction so Lutz-Nagell would yield a contradiction). Thus I will discard the question on $E(\mathbb{Q})$ and ask it instead for the torsion subgroup $E(\mathbb{Q})_{tors}$.

I'm not sure if one can use Lutz-Nagell to do a case analysis here, but it is certainly harder than using results about $E(\mathbb{Q}_p)$ - so I will do that. Firstly notice that $E(\mathbb{Q}) \subset E(\mathbb{Q}_p)$.

If $q \neq 5$, $E$ has good reduction at $p = 5$ (by your discriminant calculation - although you are missing a factor of $16$). By formal groups $E_1(\mathbb{Q}_5) \cong (\mathbb{Z}_5, +)$ which is torsion free (Silverman AEC IV 6.4 and VII 2.2). Hence $$E(\mathbb{Q})_{tors} \hookrightarrow E(\mathbb{Q}_5)_{tors} \hookrightarrow E(\mathbb{Q}_5)/E_1(\mathbb{Q}_5) \cong \tilde{E}(\mathbb{F}_5)$$

You can check that for curves of the form $(\star)$ have $6$ points. Hence $\#E(\mathbb{Q})_{tors} | 6$.

We deal with $q = 5$ separately. As above in this case $E$ has good reduction at $11$, and we have that $\#E(\mathbb{F}_{11}) = 12$ for each $t = 0, ..., 5$ - so $\#E(\mathbb{Q})_{tors} | 12$.

Aside: The $5$ and $11$ do not come out of thin air. If $E: y^2 = x^3 + k$ where $k \in \mathbb{Z} \setminus \{0\}$, can you prove that $\#E(\mathbb{F}_p) = p + 1$ for every $p \equiv 2 \pmod 3$ not dividing $2, 3$ or $k$?

Thus we only need to check whether $E$ has rational $2$ or $3$ torsion. The 2-torsion is easy, $E$ has 2-torsion if and only if $x^3 + q^t$ has a root. This is the case if and only if $t = 0, 3$, and there is at most $1$ root - hence at most $1$ 2-torsion point.

The $3$-torsion points on $E$ are the inflection points. The Hessian determinant of $X^3 + kZ^3 - Y^2Z$ is $$24X(3q^tZ^2 + Y^2)$$ hence we have affine 3-torsion points if and only if there is a point $(x, y) \in E(\mathbb{Q})$ satisfying $$24x(3q^t + y^2) = 0$$

Since $3q^t > 0$ the second factor cannot be $0$, hence $x = 0$. In that case $y^2 = q^t$, hence $E(\mathbb{Q})$ contains a $3$-torsion point if and only if $t$ is even.

Thus \begin{align*} E(\mathbb{Q})_{tors} &= \begin{cases} \{O\} &t \equiv 1, 5 \pmod{6} \\ \mathbb{Z}/2\mathbb{Z} &t \equiv 3 \pmod{6} \\ \mathbb{Z}/3\mathbb{Z} &t \equiv 2, 4 \pmod{6} \\ \mathbb{Z}/6\mathbb{Z} &t \equiv 0 \pmod{6} \end{cases} \end{align*}

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My challenge to you OP, can you generalise the above so that we do not need to assume that $k$ is a prime power?

Answer 2

The first thing i did was to write a quick code in sage to check some cases with small values of $q$ for the special value $t=5$. (It is the first "non-trivial" value for $t$ which is one modulo four.)

PRIMES = [q for q in primes(100) if q % 4 == 1]
for q in PRIMES: 
    k = q^5 
    E = EllipticCurve(QQ, [0, k]) 
    print(f"q = {q} and k = {k} = {k.factor()}")
    print('E :: ', E)
    r = E.rank(only_use_mwrank=False)
    print(f"\tE has rank {r}")
    if r > 0:
        print("\tGenerator(s): {}".format([P.xy() for P in E.gens()]))
    tor = E.torsion_order()
    print(f"\tE has torsion order {tor}")
    if tor > 1:
        print("\tTorsion points: O and {}"
              .format([P.xy() for P in E.torsion_points() if P != E(0)]))
    print('\n')

This gives:

q = 5 and k = 3125 = 5^5
E ::  Elliptic Curve defined by y^2 = x^3 + 3125 over Rational Field
    E has rank 0
    E has torsion order 1


q = 13 and k = 371293 = 13^5
E ::  Elliptic Curve defined by y^2 = x^3 + 371293 over Rational Field
    E has rank 1
    Generator(s): [(-77817/1681, -35950810/68921)]
    E has torsion order 1


q = 17 and k = 1419857 = 17^5
E ::  Elliptic Curve defined by y^2 = x^3 + 1419857 over Rational Field
    E has rank 0
    E has torsion order 1


q = 29 and k = 20511149 = 29^5
E ::  Elliptic Curve defined by y^2 = x^3 + 20511149 over Rational Field
    E has rank 1
    Generator(s): [(-18963613/85849, 78472742852/25153757)]
    E has torsion order 1


q = 37 and k = 69343957 = 37^5
E ::  Elliptic Curve defined by y^2 = x^3 + 69343957 over Rational Field
    E has rank 0
    E has torsion order 1


q = 41 and k = 115856201 = 41^5
E ::  Elliptic Curve defined by y^2 = x^3 + 115856201 over Rational Field
    E has rank 0
    E has torsion order 1


q = 53 and k = 418195493 = 53^5
E ::  Elliptic Curve defined by y^2 = x^3 + 418195493 over Rational Field
    E has rank 0
    E has torsion order 1


q = 61 and k = 844596301 = 61^5
E ::  Elliptic Curve defined by y^2 = x^3 + 844596301 over Rational Field
    E has rank 0
    E has torsion order 1


q = 73 and k = 2073071593 = 73^5
E ::  Elliptic Curve defined by y^2 = x^3 + 2073071593 over Rational Field
    E has rank 0
    E has torsion order 1


q = 89 and k = 5584059449 = 89^5
E ::  Elliptic Curve defined by y^2 = x^3 + 5584059449 over Rational Field
    E has rank 0
    E has torsion order 1


q = 97 and k = 8587340257 = 97^5
E ::  Elliptic Curve defined by y^2 = x^3 + 8587340257 over Rational Field
    E has rank 0
    E has torsion order 1

Under the results there are also curves of rank $>0$. And the torsion part is trivial in all examples. (Instead of an expected order $5+1=6$ as in the OP.)

The computer experiment is not confirming the claimed order of the torsion part of the elliptic curve. (It may be that i did not understand the question. Under the given circumstances, the second question, based on the first one, is missing the ground.)

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(There is restriction to $x$ being even in the OP, i can only understand it for the torsion points, if any. However, there are no such points in the above examples.)