How to find positive rational points for both x and y in $x^3+y^3=1141$?

Question

I did some research about finding a rational solution for this equation $$ x^3 + y^3 = 1141\ , $$ and learned a little about elliptic curves. However, those solutions require a known rational solution. How do you solve this without a known rational solution?

I am able to find $(-19,20)$ and $(20,-19)$ but I need both coordinates to be positive and rational.

Answer 1

There is a simple way to pass from the given equation to the equation of an elliptic curve in short Weierstraß form. First of all, a rational solution to the given equation $A=x^3+y^3$ is immediately leading via $x=v/u$, $y=1/u$ to one for: $$ E_{u,v}\ :\qquad Au^3 -v^3=1\ ,\qquad A=1141\text{ for short.} $$ We use the substitutions: $$ \left\{ \begin{aligned} u &= +\frac{6X}{Y+36A}\ ,\\ v &= -\frac{Y-36A}{Y+36A}\ , \end{aligned} \right. \qquad \left\{ \begin{aligned} X &= +12A\frac{u}{v+1}\ ,\\ Y &= -36A\frac{v-1}{v+1}\ , \end{aligned} \right. $$ and pass from the equation $(E_{u,v})$ to $$ E_{X,Y}\ :\qquad Y^2 = X^3 -432A^2\ . $$ For $A=1141$ the curve $E_{X,Y}$ has the following group of rational points: $$ E_{X,Y}(\Bbb Q)=\Bbb Z\cdot G\ , $$ where $G$ is the rational point $G=(13692, 1601964)$ computed below:

E = EllipticCurve(QQ, (0, -432*1141^2))
r = E.rank()
print(f'E has the torsion points: {E.torsion_points()}')
print(f'E has the rank: {r=}')
print(f'E has the generators: {[gen.xy() for gen in E.gens()]}')

... with the results:

E has the torsion points: [(0 : 1 : 0)]
E has the rank: r=1
E has the generators: [(13692, 1601964)]

So it is natural to compute some low multiples of $G$, and the corresponding points on $E_{u,v}$: $$ \scriptscriptstyle \begin{array}{|r|c|c|c|c|c|} \hline n & X(nG) & Y(nG) & u & v & \text{positive components?}\\ \hline -5 &\frac{107919499858822483698972}{130671988469474077921} & \frac{1419951486140003132497725950031444}{1493735662882618063636173450031} &\frac{90099894711425518980954732151}{764152273525981384682261984900} & \frac{729583389356636678953911465131}{764152273525981384682261984900} & OK\\\hline -4 &\frac{1505788158419692}{1557925852561} & -\frac{35882783087287412944556}{1944554753465210809} &\frac{1235415052524024021}{4819429005921295601} & \frac{12681563775265601680}{4819429005921295601} & OK\\\hline -3 &\frac{56173881}{36100} & -\frac{388326874221}{6859000} &-\frac{7115358260}{11842954469} & -\frac{74451906469}{11842954469} & \\\hline -2 &\frac{579628}{169} & -\frac{438203332}{2197} &-\frac{4953}{38120} & -\frac{57893}{38120} & \\\hline -1 &13692 & -1601964 &-\frac{1}{19} & -\frac{20}{19} & \\\hline 1 &13692 & 1601964 &\frac{1}{20} & -\frac{19}{20} & \\\hline 2 &\frac{579628}{169} & \frac{438203332}{2197} &\frac{4953}{57893} & -\frac{38120}{57893} & \\\hline 3 &\frac{56173881}{36100} & \frac{388326874221}{6859000} &\frac{7115358260}{74451906469} & -\frac{11842954469}{74451906469} & \\\hline 4 &\frac{1505788158419692}{1557925852561} & \frac{35882783087287412944556}{1944554753465210809} &\frac{1235415052524024021}{12681563775265601680} & \frac{4819429005921295601}{12681563775265601680} & OK\\\hline 5 &\frac{107919499858822483698972}{130671988469474077921} & -\frac{1419951486140003132497725950031444}{1493735662882618063636173450031} &\frac{90099894711425518980954732151}{729583389356636678953911465131} & \frac{764152273525981384682261984900}{729583389356636678953911465131} & OK\\\hline \end{array} $$ We come back to the $(x,y)$-world, and compute the corresponding solutions: $$ \scriptscriptstyle \begin{array}{|r|c|c|c|} \hline n & x(nG) & y(nG) & \text{positive components?}\\ \hline -8 &-\frac{20197700640338492383064954268566600825502944939084720003781660098370987006401}{2381305545249687781279351690485946649094912277333057869129944426443857745179} & \frac{28702892949049111292455992823667595666172140944311096279912913428625801411360}{2381305545249687781279351690485946649094912277333057869129944426443857745179} & \\\hline -7 &-\frac{10107111001800429822916755727649148165691949623610770407860}{4440038149850805694297164372288229670320006269351167798801} & \frac{46555479964811694055612989260219480802501737256206366430781}{4440038149850805694297164372288229670320006269351167798801} & \\\hline -6 &\frac{3251781815449699181565034034351186398064507}{982759974343824034008475397266016512177560} & \frac{10159490447312635405133019202725783999183493}{982759974343824034008475397266016512177560} & OK\\\hline -5 &\frac{729583389356636678953911465131}{90099894711425518980954732151} & \frac{764152273525981384682261984900}{90099894711425518980954732151} & OK\\\hline -4 &\frac{12681563775265601680}{1235415052524024021} & \frac{4819429005921295601}{1235415052524024021} & OK\\\hline -3 &\frac{74451906469}{7115358260} & -\frac{11842954469}{7115358260} & \\\hline -2 &\frac{57893}{4953} & -\frac{38120}{4953} & \\\hline -1 &20 & -19 & \\\hline 1 &-19 & 20 & \\\hline 2 &-\frac{38120}{4953} & \frac{57893}{4953} & \\\hline 3 &-\frac{11842954469}{7115358260} & \frac{74451906469}{7115358260} & \\\hline 4 &\frac{4819429005921295601}{1235415052524024021} & \frac{12681563775265601680}{1235415052524024021} & OK\\\hline 5 &\frac{764152273525981384682261984900}{90099894711425518980954732151} & \frac{729583389356636678953911465131}{90099894711425518980954732151} & OK\\\hline 6 &\frac{10159490447312635405133019202725783999183493}{982759974343824034008475397266016512177560} & \frac{3251781815449699181565034034351186398064507}{982759974343824034008475397266016512177560} & OK\\\hline 7 &\frac{46555479964811694055612989260219480802501737256206366430781}{4440038149850805694297164372288229670320006269351167798801} & -\frac{10107111001800429822916755727649148165691949623610770407860}{4440038149850805694297164372288229670320006269351167798801} & \\\hline 8 &\frac{28702892949049111292455992823667595666172140944311096279912913428625801411360}{2381305545249687781279351690485946649094912277333057869129944426443857745179} & -\frac{20197700640338492383064954268566600825502944939084720003781660098370987006401}{2381305545249687781279351690485946649094912277333057869129944426443857745179} & \\\hline \end{array} $$ Which is the "next point" with positive components on the given curve $x^3 + y^3 =A$?

As seen from the table, the symmetry $(x,y)\leftrightarrow (y,x)$ of the given curve corresponds to $nG\leftrightarrow -nG$. So we restrict to $n>0$. Then the "next point" with positive components is obtained by chance for $n=14$, and it has the components $$x=a/d^2\ , \qquad y=b/d^3\ ,$$ where:

$$ \scriptscriptstyle \begin{aligned} d &= 13 \cdot 67 \cdot 79 \cdot 229 \cdot 1973 \cdot 4273 \cdot 11801 \cdot 20369 \cdot 999912983 \cdot 7444893961 \cdot 98219876142023 \cdot 26821632623650536155443 \\ &=626230968765064500908406698002163698753726606441611774053929714399243715527427 \\[2mm] a&=366512605460575637459879550925446959825959721136405443562315244392970531278204356287931431460810492826906889647704298737769629439488179659856286740770212345068 \\ b&=3913269940586688523023028783172102841052906712137213720262934652110477241689555806926413983594233087186943792852425247864207345435773670081587851493680903057701817406317425792305980124580540094215664557964734757891690564577112872034328188 \end{aligned} $$

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Note: The above first table was produced via:

A = 1141
G = EllipticCurve(QQ, (0, -432*A^2)).gens()[0]
for n in [-5..-1] + [1..5]:
    X, Y = (n*G).xy()
    u, v = 6*X/(Y + 36*A), -(Y - 36*A)/(Y + 36*A)
    ok = 'OK' if u > 0 and v > 0 else ''
    print(f'{n} &'
          f'{latex(X)} & {latex(Y)} &'
          f'{latex(u)} & {latex(v)} & {ok}\\\\\\hline')

and the second table, with the same G, and some more $n$-values:

for n in [-8..-1] + [1..8]:
    X, Y = (n*G).xy()
    u, v = 6*X/(Y + 36*A), -(Y - 36*A)/(Y + 36*A)
    x, y = v/u, 1/u
    ok = 'OK' if x > 0 and y > 0 else ''
    print(f'{n} &'
          f'{latex(x)} & {latex(y)} & {ok}\\\\\\hline')

Answer 2

A key idea for solving $x^3 + y^3 = 1141$ is point doubling. I refer to the book by L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXI, section 'Three equal sums of two cubes', pp. $572$-$578$. On page $572$ "Numbers the sum of two rational cubes: $x^3 + y^3 = Az^3$":

$\quad$ J. Pretet$^{181}$ employed Fermat's process to get the solution $$X = x(2y^3+x^3), \qquad Y = -y(2x^3+y^3), \qquad Z = z(x^3-y^3). $$

Given one solution $\;(x:y:z) = (20:-19:1)\;$ of $\;x^3+y^3=1141z^3,\;$ its double is $(-114360: 173679: 14859)$ which is equivalent to $(38120: -57893: -4953)$ in homogeneous coordinates. In order to get both $x$ and $y$ positive you can keep doubling the points. In fact the next doubling gives $$(12681563775265601680: 4819429005921295601: 1235415052524024021).$$

All the points $(x:y:z)$ in homogeneous coordinates that satisfy the elliptic curve equation $x^3+y^3=Az^3$ form an abelian group with a geometric group law. Doubling a point is a special case of addition of points. In any particular case, there may or may not be any positive rational solutions. However, if there is a non-torsion such solution (as is the case here) then there will be infinitely many such close to it.

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In general, the solutions to $\,x^3 + y^3 = Az^3\,$ are given by generalized Somos-$5$ sequences. In the specific case of $\,A=1141\,$ we know that $\;(x:y:z) = (20:-19:1)\;$ is one solution and others can be found through doubling and addition of points. Now define the sequence of points $\,(x_n:y_n:z_n)\,$ through initial values $$ x_{-2} = 38120,\; x_{-1} = 19,\; x_0 = 1,\; x_1 = 20,\; x_2 = 57803 $$ and the recursion relation for all integer $\,n\,$ $$ x_{n+3}x_{n-2}=7115358260x_{n+2}x_{n-1}-249427630228957 x_{n+1}x_n. $$ Define the other sequence $\,y_n := -x_{-n}\,$ for all integer $\,n.\,$ Also define $\,z_n := \pm\sqrt[3]{(x_n^3 + y_n^3)/A}\,$ for $\,0\le n\le 5,\; z_{n} = -z_{-n}\,$ for $\,n<0\,$ and $\,y_n, z_n\,$ satisfy the same recursion relation as $\,x_n.\,$ You can verify that $\, x_n^3 + y_n^3 = Az_n^3\,$ for all integer $\,n.\,$

For convenience and as a check, here is a table of sequence values:

$$ \begin{array}{|r|r|r|r|} \hline n && x_n & y_n & z_n \\ \hline -3 && 11842954469 & -74451906469 & -7115358260 \\ \hline -2 && 38120 & -57893 & -4953 \\ \hline -1 && 19 & -20 & -1 \\ \hline 0 && 1 & -1 & 0 \\ \hline 1 && 20 & -19 & 1 \\ \hline 2 && 57893 & -38120 & 4953 \\ \hline 3 && 74451906469 & -11842954469 & 7115358260 \\ \hline 4 && 12681563775265601680 & 4819429005921295601 & 1235415052524024021 \\ \hline \end{array}$$

Approximately $1/3$ of all points have $\,x_n\,$ and $\,y_n\,$ of the same algebraic sign as $\,z_n\,$ and thus, the affine point $\,(x,y) := \big(\frac{x_n}{z_n}, \frac{y_n}{z_n}\big)\,$ satisfies $\,x^3+y^3 = 1141,\,$ is positive and rational since $\,x_n, y_n, z_n\in\mathbb{Z}.\,$ These points come in runs of either three or four consecutive points.

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The general situation is we want to find rational points $\,(x,y)\,$ of the curve $\,E\!: x^3 + y^3 = A\,$ where $\,A\,$ is a fixed integer. It is natural to use homogeneous coordinates to modify the equation to $\,x^3 + y^3 = Az^3\,$ where $\,(x:y:z)\,$ is the homogeneous coordinates of the point $\,(x/z,y/z).\,$ The set of all points on the curve $\,E\,$ including the point at infinity $\mathcal{O}$ forms an abelian group using elliptic curve addition $\oplus$. The negation of point $\,(x,y)\,$ is $\,\ominus(x,y) = [-1](x,y) := (y,x)\,$ or $\,(y:x:z)\,$ in homogeneous coordinates. The additive identity is $\mathcal{O}:=\,(1:-1:0).\,$ The negative of the double of a point is, as given in Dickson,

$$ [-2](x:y:z) = (x(2y^3+x^3): -y(2x^3+y^3): z(x^3-y^3)). $$

Given two points $\,P_1 = (x_1:y_1:z_1)\,$ and $\,P_2 = (x_2:y_2:z_2),\,$ the negative of the group sum is

$$ \ominus(P_1 \oplus P_2) := ( x_1 x_1 y_2 z_2 - y_1 z_1 x_2 x_2: y_1 y_1 x_2 z_2 - x_1 z_1 y_2 y_2: z_1 z_1 x_2 y_2 - x_1 y_1 z_2 z_2). $$

The group of rational points is finitely generated and may have zero or more generators with torsion points. The Wikipedia article Elliptic curve covers some of the basics. The peculiar appearance of negative of addition is explained by the result that $\,P_1 \oplus P_2 \oplus P_3 = \mathcal{O}\,$ for addition of points on the curve is equivalent to $\,u_1P_1 + u_2P_2 + u_3P_3 = 0\,$ for some constants $\,u_1,u_2,u_3\,$ for the homogeneous point triples.