# a Diophantine equation question

Jaap Spies j.spies at hccnet.nl
Sun May 27 01:04:22 CEST 2007

```N. J. A. Sloane wrote:
> Jon Schoenfield wrote today asking a question that I think Magma
> could answer - maybe someone could help?
> [Becuase these questions reduce to finding integer solutions
> to cubic equations.]
> Neil
>
>>From jonscho at hiwaay.net  Wed May  9 20:16:01 2007

>>
>> (c-1)^3 + c^3 + (c+1)^3 = 3c^3 + 6c = 3c(c^2 + 2)
>>

The equation s^2 = 3c^3 + 6c, can be transformed with the substitution
X = 3c, Y = 3s in

Y^2 = X^3 + 18X,

a form of the Weierstrass equation of an elliptic curve:

Y^2 = X^3 + aX^2 + bX + c, with a = c = 0.

The problem is finding rational (integral) points on this curve.

MAGMA can help, but also SAGE comes to the rescue.

SAGE: http://modular.math.washington.edu/sage/

We remember that the points on the elliptic curve E have a group
structure (see all over the internet).

We enter the elliptic curve in general Weierstrass form (0,0,0,18,0) or with shorthand
into the short Weierstrass form:

> sage: E = EllipticCurve([18,0])
>
> sage: E.conductor()
>  2304
>
> sage: E.torsion_subgroup()
>  Torsion Subgroup isomorphic to Multiplicative Abelian Group isomorphic to C2 associated to the Elliptic Curve defined by y^2  = x^3 + 18*x over Rational Field
>
> sage: E.rank()
>  1
>
> sage: E.is_on_curve(3,9)
>  True
>
> sage: E
>  Elliptic Curve defined by y^2  = x^3 + 18*x over Rational Field
>
> sage: E.point_search(10)
> Entering qsieve::search: y^2 = 1x^3 + 0x^2 + 18x^1 + 0
> P1 = [0:1:0]     is torsion point, order 1
> Using speed ratios 1000 and 5
> 9 primes used for first stage of sieving
> 52 primes used for both stages of sieving together.
> Sieving primes:
>  First stage: 109, 41, 13, 197, 241, 181, 113, 193, 229,
>  Second stage: 239, 223, 199, 191, 167, 151, 127, 251, 227, 211, 103, 179, 163, 79, 71, 139, 131, 173, 107, 47, 83, 67, 31, 59, 23, 157, 43, 53, 17, 233, 73, 149, 89, 137, 29, 19, 101, 97, 61, 7, 11, 37, 3,
> Probabilities: Min(109) = 0.418231, Cut1(229) = 0.495662, Cut2(3) = 0.777778, Max(5) = 1
>
> Forbidden divisors of the denominator:
>
>
> Try to find the points up to height 22026
> x = 0/1 gives a rational point.
> P1 = [0:0:1]     is torsion point, order 2
> x = 3/1 gives a rational point.
> P1 = [3:9:1]      is generator number 1
> saturating up to 20...Checking 2-saturation
> Points have successfully been 2-saturated (max q used = 7)
> Checking 3-saturation
> Points have successfully been 3-saturated (max q used = 11)
> Checking 5-saturation
> Points have successfully been 5-saturated (max q used = 13)
> Checking 7-saturation
> Points have successfully been 7-saturated (max q used = 83)
> Checking 11-saturation
> Points have successfully been 11-saturated (max q used = 43)
> Checking 13-saturation
> Points have successfully been 13-saturated (max q used = 103)
> Checking 17-saturation
> Points have successfully been 17-saturated (max q used = 29)
> Checking 19-saturation
> Points have successfully been 19-saturated (max q used = 151)
> done
> x = 6/1 gives a rational point.
> P2 = [6:18:1]    = -1*P1 (mod torsion)
> x = 72/1 gives a rational point.
> P2 = [72:612:1]  = 2*P1 (mod torsion)
> x = 1/4 gives a rational point.
> P2 = [2:17:8]    = -2*P1 (mod torsion)
> x = 1587/121 gives a rational point.
> P2 = [17457:66447:1331]  = -3*P1 (mod torsion)
> x = 726/529 gives a rational point.
> P2 = [16698:63558:12167]         = 3*P1 (mod torsion)
>  [(3 : 9 : 1)]
>
> sage:
>

What does this mean? The group is finitely generated. There is one
generator point P1 = (3,9) in affine coordinates. The points (6,18)
and (72,612) are the only other integral points on the curve.
Corresponding to c = 1, 2, 24 in the original equation.

>>sequence's terms, 0, 9, 36, and 41616, correspond to c = 0, 1, 2, and 24.
>>After an exhaustive search up through c = 4*10^8, I can say that the next
>>term exceeds 1.92*10^26 -- if it exists at all, which I doubt, but can't
>>prove ....
>>
>>

Hope this helps.

Jaap

```