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I have tried some case about this question such as $n=9$.

And I find out when $n=9$, $X^3+Y^3=9$ has infinite rational solutions. This can be proved by using elliptic curves and geometric constructions. We can find tangent line of the initial solution and then determine if this tangent line intersect with the curve or not.

But this method can only use when you know the value of n I think. I have no idea how to find out all the value of $n$ which satisfy the requirement.

Any hint or solution is appreciated. Thanks.


Actually there is a previous question for this question:

"Prove that the torsion points of elliptic curve over $\mathbb{Q}$ are finitely many."

My solution for this question is "By Mordell-Weil thoerem, $E(\mathbb{Q})$ is a finitely generated abelian group, thus by the structure theorem of finitely generated abelian groups, the torsion part of $E(\mathbb{Q})$ is a finite group."

Our teacher gave me this comment, "You can use the technique of Mordell-Weil thoerem to prove that rational torsion points are finite, don't apply Mordell-Weil thoerem directly. Proving that torsion points are finite is far easier than the Mordell-Weil thoerem." But I have no idea how to use the technique of Mordell-Weil thoerem to prove that rational torsion points are finite.

Helping on this question or the question about find all $n$ is appreciated. Thanks.


I got a hint from my teacher but I still cannot figure out. So I decide to post the hint here.

Hint: The general problem is this: given a rational elliptic curve of "family", how to find all elliptic curves of this family with rank 0 and non-trivial torsion points?

Guess: there are only a limited number of good families, and there are effective algorithms.

Question: what is a good family?

Case: $Y^3+Z^3=n$

What is the method for this example?

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  • 3
    $\begingroup$ Do you mean finite amount of rational solutions? $\endgroup$
    – Jakobian
    Commented Aug 5, 2021 at 5:41
  • $\begingroup$ Yes, and $n \in \mathbb{Q}$. $\endgroup$
    – visiteur
    Commented Aug 5, 2021 at 5:46

2 Answers 2

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Let $a, b, c$ be non-zero rational numbers. Then the genus 1 surface $C: ax^3 + by^3 + c=0$ has infinitely many non-zero rational points if and only if the Mordel-Weil group of the elliptic curve $Y^2 = X^3 + 432abc$ has rank $> 0$.

In other words, if $C$ has a non-zero rational point, one can derive infinitely many other non-zero rational points.

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  • $\begingroup$ Yes. But I cannot see the requirement of $n$. Can you explain a little more explicit? $\endgroup$
    – visiteur
    Commented Aug 5, 2021 at 8:41
  • $\begingroup$ Since we don’t have $Y^2$ $\endgroup$
    – visiteur
    Commented Aug 5, 2021 at 8:43
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The question seems asks for a "pattern" for all $n$, such that there are only finitely many rational solution of the (affine) equation: $$ \tag{$*$} X^3+ Y^3 = n\ . $$ (Keywords: Sylvester's conjecture.)

This answer cites some results, and shows that even in the special case of a prime $n$ things are nontrivial. Using sage two tables are plotted, one for the first few integers $n\ge 1$, one for the case $n=p$ prime congruent to $1$ modulo nine.


There are some articles written on this topic:

This papers have a special research target, consider the case when $n$ is directly related to a prime $p$ (with further restrictions), e.g. $n=p, p^2, 2p, 2p^2, 25p^2$. The introductions are informative for the present question. The situation for prime values of $n$ is the common research theme. This answer merely uses computer aid, sage, to support the results.

I am citing now from the paper on mock Heegner points, (D,V,1)...

Conjecture: (Sylvester, Selmer) Let $E_n$ be the elliptic curve with (affine) equation $X^3+Y^3=n$. If $n=p$ is a prime, and $p=4,7,8$ modulo nine, then the rank of $E_p(\Bbb Q)$ is positive.

An explicit $3$-descent argument by Satgé shows:

$$\operatorname{rank}E_p(\Bbb Q)\le\begin{cases} 0 &\text{ for }p\equiv 2, 5&&\text{ modulo nine,}\\ 1 &\text{ for }p\equiv 4, 7,8&&\text{ modulo nine,}\\ 2 &\text{ for }p\equiv 1&&\text{ modulo nine,} \end{cases}$$

(This covers only the case of a prime $n$.)

The case $p=1$ modulo nine shows that from situation to situation, the special arithmetic of the corresponding elliptic curve determines if we have rank $0$ or rank $2$. (The sign of the functional equation for $L(E_p,\cdot)$ is plus in this case.)

In this answer there is only some computer algebra evidence for the first few natural values of $n$ up to $200$. For such values of $n$ the code in the sequel realizes the following "plan". It starts with the curve $E_n$ with equation $X^3+Y^3=n$, brings it into the Weierstraß form $E'_n$ with equation $y^2=x^3-432n^2$, then asks for the rank and the generators of $E'(\Bbb Q)$, and - if any - brings them back to $E_n(\Bbb Q)$. The data is collected in a string that is suitable to be copy-pasted here inside an array block.


The passage from $E_n$ to $E'_n$ is as follows: $$ \begin{aligned} X &=\frac{36n-y}{6x}\ , & Y&=\frac{36n+y}{6x}\ ,\\[2mm] x &=\frac{12n}{X+Y}\ ,& y &=-36n\cdot\frac{X-Y}{X+Y}\ . \end{aligned} $$


Let us check that the above birational maps $(x,y)\to(X,Y)$ and $(X,Y)\to(x,y)$ are well defined, and inverse to each other. (Please skip if this feels too explicit.)

  • We start with $(x,y)$ satisfying $y^2=x^3-432n^2$. Then: $$ \begin{aligned} X^3+Y^3&=\frac 1{216x^3}((36n-y)^3+(36n+y)^3 \\ &=\frac 2{216x^3}((36n)^3 +3(36n)y^2) \\ &=\frac2{216(432n^2+y^2)}\cdot (36n)\cdot ((36n)^2+3y^2) \\ &=n\ . \end{aligned} $$
  • Assume now $X^3+Y^3=n$, then: $$ \begin{aligned} x^3-y^2 &=(12n)^3\cdot \frac {(X+Y)^3} - (36n)^2\frac {(X-Y)^2}{(X+Y)^2} \\ &=(12n)^2\cdot \frac 1{(X+Y)^3}(12n -9(X-Y)^2(X+Y)) \\ &=(12n)^2\cdot 3\cdot \frac {(X^3 + Y^3)+3XY(X+Y)}{4n-3(X^3 + Y^3)+3XY(X+Y)} \\ &=(12n)^2\cdot 3\cdot \frac {n+3XY(X+Y)}{n+3XY(X+Y)} \\ &=(12n)^2\cdot 3=432n^2\ . \end{aligned} $$
  • It remains to check that the formulas for the passage between the $(X,Y)$-world and the $(x,y)$-world are inverse to each other. We start with the formulas for $(X,Y)$ and compute successively $\displaystyle X+Y=2\cdot \frac{36n}{6x}=\frac{12n}x$, $\displaystyle X-Y=-2\cdot \frac{y}{6x}=\frac{12n}x$, $\displaystyle \frac{12n}{X+Y}=x$, and $\displaystyle -(36n)\cdot{X-Y}{X+Y}= -(36n)\cdot\frac{-2\cdot y/(6x)}{-2\cdot(36n)/(6x)}=y $. The other direction is similar.

The code implementing the "plan" above uses the above formulas to pass from the Selmer curve $E_n$ to the elliptic curve in Weierstraß form $E'_n$. (And computes rank and generators in this world, the world where sage implements them.)

s = ''    # latex string to be inserted in an array block

def get_selmer_point(n, P):
    """Here P = (x, y) is a point on y² = x³ - 432n²"""
    x, y = P.xy()
    X, Y = (36*n - y)/6/x, (36*n + y)/6/x
    denom = gcd(X.denominator(), Y.denominator())
    return (X*denom, Y*denom, denom)

for n in [1..120]:
    print(f'Computations for n = {n}')
    E = EllipticCurve(QQ, [0, -432*n^2])
    print(f'E = {E}')
    try:
        r = E.rank(only_use_mwrank=False)
        print(f'\tRANK = {r}')
        gens = [get_selmer_point(n, P) for P in E.gens()]
        gens_info = '\\ ,\\ '.join([str(Q) for Q in gens])
    except:
       print(f'*** rank computation error, using analytic rank... ***')
       r = E.analytic_rank()
       gens_info = '?' if r > 0 else ''
    
    prime_info = '\\text{PRIME}' if n.is_prime() else ''
    s += f'{n} & {n%9} & {prime_info} & {r} & {gens_info}\\\\\\hline\n'

print(s)

The results are presented in the following latex array block: $$ \begin{array}{|r||r|l|c|l|} \hline n & n[9] & \text{prime?} & r & \text{generators $(X,Y,Z)$ with }X^3+Y^3=nZ^3\\\hline\hline 1 & 1 & & 0 & \\\hline 2 & 2 & \text{PRIME} & 0 & \\\hline 3 & 3 & \text{PRIME} & 0 & \\\hline 4 & 4 & & 0 & \\\hline 5 & 5 & \text{PRIME} & 0 & \\\hline 6 & 6 & & 1 & (17, 37, 21)\\\hline 7 & 7 & \text{PRIME} & 1 & (-1, 2, 1)\\\hline 8 & 8 & & 0 & \\\hline 9 & 0 & & 1 & (1, 2, 1)\\\hline 10 & 1 & & 0 & \\\hline 11 & 2 & \text{PRIME} & 0 & \\\hline 12 & 3 & & 1 & (19, 89, 39)\\\hline 13 & 4 & \text{PRIME} & 1 & (2, 7, 3)\\\hline 14 & 5 & & 0 & \\\hline 15 & 6 & & 1 & (397, 683, 294)\\\hline 16 & 7 & & 0 & \\\hline 17 & 8 & \text{PRIME} & 1 & (-1, 18, 7)\\\hline 18 & 0 & & 0 & \\\hline 19 & 1 & \text{PRIME} & 2 & (3, 5, 2)\ ,\ (1, 8, 3)\\\hline 20 & 2 & & 1 & (1, 19, 7)\\\hline 21 & 3 & & 0 & \\\hline 22 & 4 & & 1 & (17299, 25469, 9954)\\\hline 23 & 5 & \text{PRIME} & 0 & \\\hline 24 & 6 & & 0 & \\\hline 25 & 7 & & 0 & \\\hline 26 & 8 & & 1 & (-1, 3, 1)\\\hline 27 & 0 & & 0 & \\\hline 28 & 1 & & 1 & (1, 3, 1)\\\hline 29 & 2 & \text{PRIME} & 0 & \\\hline 30 & 3 & & 2 & (107, 163, 57)\ ,\ (-19, 289, 93)\\\hline 31 & 4 & \text{PRIME} & 1 & (-65, 137, 42)\\\hline 32 & 5 & & 0 & \\\hline 33 & 6 & & 1 & (523, 1853, 582)\\\hline 34 & 7 & & 1 & (-359, 631, 182)\\\hline 35 & 8 & & 1 & (2, 3, 1)\\\hline 36 & 0 & & 0 & \\\hline 37 & 1 & \text{PRIME} & 2 & (18, 19, 7)\ ,\ (-1, 10, 3)\\\hline 38 & 2 & & 0 & \\\hline 39 & 3 & & 0 & \\\hline 40 & 4 & & 0 & \\\hline 41 & 5 & \text{PRIME} & 0 & \\\hline 42 & 6 & & 1 & (-71, 449, 129)\\\hline 43 & 7 & \text{PRIME} & 1 & (1, 7, 2)\\\hline 44 & 8 & & 0 & \\\hline 45 & 0 & & 0 & \\\hline 46 & 1 & & 0 & \\\hline 47 & 2 & \text{PRIME} & 0 & \\\hline 48 & 3 & & 1 & (34, 74, 21)\\\hline 49 & 4 & & 1 & (-2, 11, 3)\\\hline 50 & 5 & & 1 & (-11267, 23417, 6111)\\\hline 51 & 6 & & 1 & (62641, 730511, 197028)\\\hline 52 & 7 & & 0 & \\\hline 53 & 8 & \text{PRIME} & 1 & (-1819, 1872, 217)\\\hline 54 & 0 & & 0 & \\\hline 55 & 1 & & 0 & \\\hline 56 & 2 & & 1 & (-2, 4, 1)\\\hline 57 & 3 & & 0 & \\\hline 58 & 4 & & 1 & (-14653, 28747, 7083)\\\hline 59 & 5 & \text{PRIME} & 0 & \\\hline 60 & 6 & & 0 & \\\hline 61 & 7 & \text{PRIME} & 1 & (-4, 5, 1)\\\hline 62 & 8 & & 1 & (7, 11, 3)\\\hline 63 & 0 & & 1 & (-1, 4, 1)\\\hline 64 & 1 & & 0 & \\\hline 65 & 2 & & 2 & (197, 323, 86)\ ,\ (1, 4, 1)\\\hline 66 & 3 & & 0 & \\\hline 67 & 4 & \text{PRIME} & 1 & (1208, 5353, 1323)\\\hline 68 & 5 & & 1 & (-472663, 2538163, 620505)\\\hline 69 & 6 & & 1 & (-10441, 15409, 3318)\\\hline 70 & 7 & & 1 & (17, 53, 13)\\\hline 71 & 8 & \text{PRIME} & 1 & (-126, 197, 43)\\\hline 72 & 0 & & 1 & (2, 4, 1)\\\hline 73 & 1 & \text{PRIME} & 0 & \\\hline 74 & 2 & & 0 & \\\hline 75 & 3 & & 1 & (-11951, 17351, 3606)\\\hline 76 & 4 & & 0 & \\\hline 77 & 5 & & 0 & \\\hline 78 & 6 & & 1 & (53, 5563, 1302)\\\hline 79 & 7 & \text{PRIME} & 1 & (-4, 13, 3)\\\hline 80 & 8 & & 0 & \\\hline 81 & 0 & & 0 & \\\hline 82 & 1 & & 0 & \\\hline 83 & 2 & \text{PRIME} & 0 & \\\hline 84 & 3 & & 1 & (323, 433, 111)\\\hline 85 & 4 & & 1 & (-2404889, 2570129, 330498)\\\hline 86 & 5 & & 2 & (60877, 106307, 25506)\ ,\ (5, 13, 3)\\\hline 87 & 6 & & 1 & (1176498611, -907929611, 216266610)\\\hline 88 & 7 & & 0 & \\\hline 89 & 8 & \text{PRIME} & 1 & (36, 53, 13)\\\hline 90 & 0 & & 1 & (-431, 1241, 273)\\\hline 91 & 1 & & 2 & (3, 4, 1)\ ,\ (23, 94, 21)\\\hline 92 & 2 & & 1 & (-3547, 25903, 5733)\\\hline 93 & 3 & & 0 & \\\hline 94 & 4 & & 1 & (15642626656646177, -15616184186396177, 590736058375050)\\\hline 95 & 5 & & 0 & \\\hline 96 & 6 & & 1 & (38, 178, 39)\\\hline 97 & 7 & \text{PRIME} & 1 & (-5, 14, 3)\\\hline 98 & 8 & & 1 & (-3, 5, 1)\\\hline 99 & 0 & & 0 & \\\hline 100 & 1 & & 0 & \\\hline 101 & 2 & \text{PRIME} & 0 & \\\hline 102 & 3 & & 0 & \\\hline 103 & 4 & \text{PRIME} & 1 & (-349, 592, 117)\\\hline 104 & 5 & & 1 & (4, 14, 3)\\\hline 105 & 6 & & 1 & (3527, 4033, 1014)\\\hline 106 & 7 & & 1 & (-140131, 165889, 25767)\\\hline 107 & 8 & \text{PRIME} & 1 & (17, 90, 19)\\\hline 108 & 0 & & 0 & \\\hline 109 & 1 & \text{PRIME} & 0 & \\\hline 110 & 2 & & 2 & (251, 629, 134)\ ,\ (-71, 181, 37)\\\hline 111 & 3 & & 0 & \\\hline 112 & 4 & & 0 & \\\hline 113 & 5 & \text{PRIME} & 0 & \\\hline 114 & 6 & & 1 & (-901, 9109, 1878)\\\hline 115 & 7 & & 1 & (-2741617, 5266097, 1029364)\\\hline 116 & 8 & & 0 & \\\hline 117 & 0 & & 1 & (-2, 5, 1)\\\hline 118 & 1 & & 0 & \\\hline 119 & 2 & & 0 & \\\hline 120 & 3 & & 1 & (397, 683, 147)\\\hline 121 & 4 & & 0 & \\\hline 122 & 5 & & 0 & \\\hline 123 & 6 & & 1 & (184223499139, 10183412861, 37045412880)\\\hline 124 & 7 & & 2 & (-1, 5, 1)\ ,\ (-443, 479, 57)\\\hline 125 & 8 & & 0 & \\\hline 126 & 0 & & 2 & (1, 5, 1)\ ,\ (-121, 127, 13)\\\hline 127 & 1 & \text{PRIME} & 2 & (-251, 378, 67)\ ,\ (-6, 7, 1)\\\hline 128 & 2 & & 0 & \\\hline 129 & 3 & & 0 & \\\hline 130 & 4 & & 1 & (52954777, 33728183, 11285694)\\\hline 131 & 5 & \text{PRIME} & 0 & \\\hline 132 & 6 & & 2 & (-901, 2089, 399)\ ,\ (-29503, 39007, 6342)\\\hline 133 & 7 & & 1 & (2, 5, 1)\\\hline 134 & 8 & & 1 & (7, 9, 2)\\\hline 135 & 0 & & 0 & \\\hline 136 & 1 & & 1 & (-2, 36, 7)\\\hline 137 & 2 & \text{PRIME} & 0 & \\\hline 138 & 3 & & 0 & \\\hline 139 & 4 & \text{PRIME} & 1 & (-7, 16, 3)\\\hline 140 & 5 & & 1 & (6623, 27397, 5301)\\\hline 141 & 6 & & 1 & (53579249, -52310249, 4230030)\\\hline 142 & 7 & & 1 & (2454839, 1858411, 530595)\\\hline 143 & 8 & & 1 & (15, 73, 14)\\\hline 144 & 0 & & 0 & \\\hline 145 & 1 & & 0 & \\\hline 146 & 2 & & 0 & \\\hline 147 & 3 & & 0 & \\\hline 148 & 4 & & 0 & \\\hline 149 & 5 & \text{PRIME} & 0 & \\\hline 150 & 6 & & 0 & \\\hline \end{array} $$ There is no predictible "simple rule" to see when the rank is zero!


Here comes a further table, containing only the cases of a prime $n$, of the shape $n=p$ with $p$ congruent to $1$ modulo $9$. Citing from the paper on mock Heegner points, (D,V,1) ...

... for $p\equiv 1$ (mod $9$), the Birch–Swinnerton-Dyer (BSD) conjecture predicts that $\operatorname{rank}E_p(\Bbb Q) = 0$ or $2$, depending on $p$ in a nontrivial way. This case was investigated by Rodriguez-Villegas and Zagier [10]: they give three methods to determine for a given prime $p$ whether or not $\operatorname{rank}E_p(\Bbb Q) = 0$.

Which is this non-trivial way? We have to look into (RV,Z). For the rank equal to zero, as asked in the OP, we expect a non-vanishing of $$ L(E'_p,1)= \frac {\sqrt 3\; \Gamma\left(\frac 1p\right)^3}{2\pi\; p^{1/3}}\cdot S_p\ ,\qquad S_p=\operatorname{Trace}(\alpha_p)=\operatorname{Trace}^2(\beta_P)\in\Bbb Z \ , $$ where $S_p$ is written alternatively as first two powers of the the trace of some algebraic numbers $\alpha_p,\beta_P$ of high degree (increasing with $p$), see loc. cit. for their definition. In §5 of (RV,Z) it is shown that $|S_p|<p/2$, so it is enought to know $S_p$ modulo $p$, and moreover, the vanishing of $S_p$ becomes equivalent to its vanishing modulo $p$. And we have modulo $p$ $$ S_p\equiv \text{(known factor)}\cdot A_{2(p-1)/9}\equiv \text{(known factor)}\cdot B^2_{2(p-1)/9}\ , $$ where the known factor above does not vanish, and the integer coefficients $A_k=B_k^2$ are extracted from the series representations: $$ \begin{aligned} {}_2F_1\left(\frac 13,\frac 13\ ;\ \frac 23\ ;\ x\right) &=\sum_{k\ge 0}\frac {A_k}{(3k)!}\; T^k\ ,\\ (1-x)^{1/24} {}_2F_1\left(\frac 13,\frac 13\ ;\ \frac 23\ ;\ x\right)^{1/2} &=\sum_{k\ge 0}\frac {B_k}{(3k)!}\; \left(\frac{-T}2\right)^k\ ,\\ &\qquad\text{ where }\\ T &= x\cdot \frac {{}_2F_1\left(\frac 23,\frac 23\ ;\ \frac 43\ ;\ x\right)^3} {{}_2F_1\left(\frac 13,\frac 13\ ;\ \frac 23\ ;\ x\right)^3} \end{aligned} $$ So for the case $n=p$ prime (and $n=p^2$) in the "complicated" case of $p\equiv 1$ modulo nine, (RV,Z) provides an answer. (Compute the corresponding coefficient $A_k=B_k^2$, take it modulo $p$, see if this residue does not vanish.)

I am trying to compute some relevant data in a table. For the first few such primes $p\equiv 1[9]$ we have in the table the rank, computed individually, together with the data for $S_p$ from the above formulas.

def A(k, p):
    R.<t> = PolynomialRing(GF(p))
    f, F, n = R(1), t^2, 1    # F is f_n(t) with notations from (RV,Z)
    while n < 3*k:
        f, F, = F, (1-t^3)*diff(F, t) + (2*n+1)*t^2*F - n^2*t*f
        n += 1
    return F(0)

def S(p):
    if not p.is_prime():    return
    k = (p - 1)/9
    if k not in ZZ:    return
    k = ZZ(k)
    F = GF(p)
    Sp = A(2*k, p) * F(-3)^(3*k -3) * prod([F(j) for j in [1..3*k]])^2
    return Sp.lift_centered()

s = ''
for p in primes(3000):
    if p % 9 != 1:    continue
    print(f'Computations for p = {p}')
    E = EllipticCurve(QQ, [0, -432*p^2])
    print(f'E = {E}')
    try:
        r = E.rank(only_use_mwrank=False)
        print(f'\tRANK = {r}')
        gens = [get_selmer_point(p, P) for P in E.gens()]
        gens_info = '\\ ,\\ '.join([str(Q) for Q in gens])
    except:
       print(f'*** rank computation error, using analytic rank... ***')
       r = E.analytic_rank()
       gens_info = '?' if r > 0 else ''
    s += f'{p} & {r} & {gens_info} & {S(p)}\\\\\\hline\n'
print(s)

Results:

$$ \begin{array}{|r||r|l||c|} \hline n & r & \text{generators $(X,Y,Z)$ with }X^3+Y^3=nZ^3 & S_p\\\hline\hline 19 & 2 & (3, 5, 2)\ ,\ (1, 8, 3) & 0\\\hline 37 & 2 & (18, 19, 7)\ ,\ (-1, 10, 3) & 0\\\hline 73 & 0 & & 1\\\hline 109 & 0 & & 1\\\hline 127 & 2 & (-251, 378, 67)\ ,\ (-6, 7, 1) & 0\\\hline 163 & 2 & (73, 90, 19)\ ,\ (-3, 11, 2) & 0\\\hline 181 & 0 & & 1\\\hline 199 & 0 & & 1\\\hline 271 & 2 & (-216, 487, 73)\ ,\ (-9, 10, 1) & 0\\\hline 307 & 0 & & 1\\\hline 379 & 2 & (-7, 15, 2)\ ,\ (-1007, 1386, 163) & 0\\\hline 397 & 2 & (37, 360, 49)\ ,\ (-11, 12, 1) & 0\\\hline 433 & 2 & (35, 37, 6)\ ,\ (181, 252, 37) & 0\\\hline 487 & 0 & & 1\\\hline 523 & 2 & (-269, 792, 97)\ ,\ (-9, 17, 2) & 0\\\hline 541 & 0 & & 1\\\hline 577 & 0 & & 1\\\hline 613 & 0 & & 1\\\hline 631 & 2 & (-3761, 4886, 465)\ ,\ (-14, 15, 1) & 0\\\hline 739 & 0 & & 4\\\hline 757 & 0 & & 1\\\hline 811 & 0 & & 1\\\hline 829 & 2 & (393, 607, 70)\ ,\ (-2591, 3420, 301) & 0\\\hline 883 & 2 & (-1482390, 5820569, 603347)\ ,\ (-13, 21, 2) & 0\\\hline 919 & 2 & (216, 703, 73)\ ,\ (-17, 18, 1) & 0\\\hline 937 & 2 & (-26, 35, 3)\ ,\ (939188250, -799690501, 69681803) & 0\\\hline 991 & 0 & & 4\\\hline 1009 & 0 & & 1\\\hline 1063 & 2 & (-28, 37, 3)\ ,\ (-22247, 23310, 1159) & 0\\\hline 1117 & 0 & & 1\\\hline 1153 & 0 & & 1\\\hline 1171 & 2 & (25, 39, 4)\ ,\ (-233, 1404, 133) & 0\\\hline 1279 & 0 & & 1\\\hline 1297 & 0 & & 1\\\hline 1423 & 0 & & 1\\\hline 1459 & 2 & (48014927, 19750897, 4329444)\ ,\ (-10440, 11899, 721) & 0\\\hline 1531 & 2 & (19, 45, 4)\ ,\ (-5309, 6840, 481) & 0\\\hline 1549 & 0 & & 1\\\hline 1567 & 2 & (307, 1260, 109)\ ,\ (-35, 44, 3) & 0\\\hline 1621 & 2 & (631, 990, 91)\ ,\ (-58445, 65006, 3591) & 0\\\hline 1657 & 2 & (1, 71, 6)\ ,\ (-1530, 3187, 259) & 0\\\hline 1693 & 0 & & 1\\\hline 1747 & 0 & & 4\\\hline 1783 & 0 & & 1\\\hline 1801 & 2 & (594, 1207, 103)\ ,\ (-1, 73, 6) & 0\\\hline 1873 & 0 & & 4\\\hline 1999 & 0 & & 4\\\hline 2017 & 0 & & 1\\\hline 2053 & 0 & & 1\\\hline 2089 & 2 & (13938, 28937, 2345)\ ,\ (396, 1693, 133) & 0\\\hline 2143 & 0 & & 1\\\hline 2161 & 0 & & 1\\\hline 2179 & 0 & & 1\\\hline 2251 & 0 & & 4\\\hline 2269 & 2 & (-2771, 5040, 361)\ ,\ (-27, 28, 1) & 0\\\hline 2287 & 0 & & 1\\\hline 2341 & 0 & & 4\\\hline 2377 & 0 & & 4\\\hline 2467 & 0 & & 4\\\hline 2503 & 0 & & 1\\\hline 2521 & 0 & & 1\\\hline 2539 & 2 & (-25, 33, 2)\ ,\ (-9467, 12006, 703) & 0\\\hline 2557 & 2 & (-3762, 6319, 427)\ ,\ (-46, 55, 3) & 0\\\hline 2593 & 0 & & 4\\\hline 2647 & 2 & (1009, 1638, 127)\ ,\ (-12740, 19599, 1273) & 0\\\hline 2683 & 0 & & 1\\\hline 2719 & 0 & & 1\\\hline 2791 & 2 & (-30, 31, 1)\ ,\ (-296496, 299287, 6433) & 0\\\hline 2917 & 0 & & 1\\\hline 2953 & 0 & & 1\\\hline 2971 & 0 & & 4\\\hline \end{array} $$ And indeed, we have solutions (and explicit generators above), iff the $S_p$ vanishes. (The computed values for $S_p$ are among $0^2, 1^2, 2^2$.)

$\endgroup$

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