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Cubics: Mordell, three cubes, and Thue

At degree three the elementary tricks run out and Diophantine solving becomes algebraic number theory in earnest. This page covers three landmark families: integral points on Mordell curves \(y^2 = x^3 + k\), the sum of three cubes \(x^3 + y^3 + z^3 = k\), and Thue equations \(F(x, y) = m\) for an irreducible binary form \(F\) of degree \(\ge 3\). Each has finitely many solutions, and each is decided by a genuine theorem rather than a scan.

In[1]:= Solve[y^2 == x^3 - 2, {x, y}, Integers]
Out[1]= {{x -> 3, y -> -5}, {x -> 3, y -> 5}}

Fermat claimed, and Euler proved, that \(y^2 = x^3 - 2\) has only the solutions \((3, \pm 5)\) — \(27 - 2 = 25\). Mathilda reproduces the theorem.

1. Mordell curves \(y^2 = x^3 + k\)

For \(k < 0\) with \(|k|\) squarefree and a class-number condition met, the equation factors in the quadratic ring \(\mathbb{Z}[\sqrt{k}]\): writing \(y^2 - k = x^3\) as \((y - \sqrt{k})(y + \sqrt{k}) = x^3\) and using unique factorisation of ideals forces \(y \pm \sqrt{k}\) to be a cube, which pins down \(x\) and \(y\) completely. The result is the complete integral-point set, and a failed descent is a proof of emptiness:

In[1]:= Solve[y^2 == x^3 - 5, {x, y}, Integers]
Out[1]= {}

In[2]:= Solve[y^2 == x^3 - 13, {x, y}, Integers]
Out[2]= {{x -> 17, y -> -70}, {x -> 17, y -> 70}}

\(y^2 = x^3 - 5\) has no integer points — the {} is the descent argument in \(\mathbb{Z}[\sqrt{-5}]\), not an exhausted search — whereas \(y^2 = x^3 - 13\) has exactly \((17, \pm 70)\), since \(17^3 - 13 = 4913 - 13 = 4900 = 70^2\).

References

  1. L. J. Mordell, Diophantine Equations, Academic Press, 1969 — the Mordell curve and the finiteness of integral points.
  2. J. Gebel, A. Pethő, H. G. Zimmer, Computing integral points on elliptic curves, Acta Arith. 68 (1994), 171–192.

2. Sum of three cubes \(x^3 + y^3 + z^3 = k\)

The three-cubes problem is famously delicate: some \(k\) have no representation at all, while others have one whose entries are astronomically large. Mathilda attacks it from both ends — a global impossibility proof, and an efficient bounded search.

2.1 A global obstruction: the mod-9 impossibility

Every cube is congruent to \(-1\), \(0\), or \(1\) modulo \(9\) (check \(0^3,\dots,8^3\)). A sum of three cubes is therefore a sum of three values from \(\{-1, 0, 1\}\), which lands in \(\{-3,\dots,3\} \equiv \{0,1,2,3,6,7,8\} \pmod 9\) and can never be \(\equiv 4\) or \(5 \pmod 9\). So \(x^3 + y^3 + z^3 = k\) with \(k \equiv \pm 4 \pmod 9\) has no integer solution whatsoever — and Mathilda returns that as a proof, with no bound required:

In[1]:= Solve[x^3 + y^3 + z^3 == 4, {x, y, z}, Integers]
Out[1]= {}

The empty set here is the congruence argument, not an exhausted search: \(4 \equiv 4 \pmod 9\), so no triple of cubes can reach it. The same holds for every \(k \equiv 4, 5 \pmod 9\) (and, since \(-k^3\) is also a cube residue, for the signed forms \(x^3 + y^3 - z^3\) too).

For a reachable residue, there is no shortcut to finding the solutions, and Mathilda uses Booker's method: for each small value of the divisor \(d = |x + y|\), solve the cube-root congruence \(z^3 \equiv k \pmod d\) and recover the candidates — far more efficient over a box than a triple loop.

In[1]:= Solve[x^3 + y^3 + z^3 == 3 && Abs[x] < 8000 && Abs[y] < 8000 && Abs[z] < 8000, {x, y, z}, Integers]
Out[1]= {{x -> -5, y -> 4, z -> 4}, {x -> 1, y -> 1, z -> 1}, {x -> 4, y -> -5, z -> 4}, {x -> 4, y -> 4, z -> -5}}

Within the box, \(k = 3\) has the well-known \((1,1,1)\) and the \((4,4,-5)\) family. This is a bounded search, so here an empty answer is a proof about the box, not the whole of \(\mathbb{Z}\):

In[2]:= Solve[x^3 + y^3 + z^3 == 42 && Abs[x] < 8000 && Abs[y] < 8000 && Abs[z] < 8000, {x, y, z}, Integers]
Out[2]= {}

There is no representation of \(42\) with all coordinates below \(8000\) — the smallest, found by Booker and Sutherland in 2019, has 17-digit entries. Because this search is intrinsically bounded, an unbounded query for a reachable residue (one the mod-9 proof cannot dismiss) is declined rather than answered:

In[3]:= Solve[x^3 + y^3 + z^3 == 3, {x, y, z}, Integers]
Out[3]= Solve[x^3 + y^3 + z^3 == 3, {x, y, z}, Integers]

\(k = 3\) is \(\equiv 3 \pmod 9\), so no congruence rules it out, and without a bound Mathilda has no complete procedure — it returns the input unchanged, an honest gap exactly as the second guarantee promises, rather than a possibly-incomplete list.

References

  1. A. R. Booker, Cracking the problem with 33, Research in Number Theory 5 (2019), 26.
  2. A. R. Booker & A. V. Sutherland, On a question of Mordell, PNAS 118 (2021), e2022377118.

3. Thue equations \(F(x, y) = m\)

3.1 The Tzanakis–de Weger method

A Thue equation sets an irreducible homogeneous binary form of degree \(\ge 3\) equal to a constant, e.g. \(x^3 - 2y^3 = 1\). Axel Thue proved in 1909 that such an equation has only finitely many solutions, but his proof was ineffective. The Tzanakis–de Weger algorithm makes it effective:

  1. work in the number field \(K = \mathbb{Q}(\theta)\), where \(\theta\) is a root of \(F(x, 1)\), and factor \(F(x, y) = a_0 \prod_i (x - \theta_i y)\);
  2. reduce the equation to unit equations in the ring of integers \(O_K\), parameterising \(x - \theta y\) by the fundamental units;
  3. bound \(\max(|x|, |y|)\) with Baker's theory of linear forms in logarithms — an astronomically large but explicit bound;
  4. shrink that bound to a searchable range by LLL lattice reduction;
  5. enumerate the reduced range and verify.

Mathilda implements the whole chain (src/solve/solvethue.c, with a real number-field layer in src/numbertheory/), so an unbounded Thue equation returns its complete, finite solution set:

In[1]:= Solve[x^3 - 2 y^3 == 1, {x, y}, Integers]
Out[1]= {{x -> -1, y -> -1}, {x -> 1, y -> 0}}

In[2]:= Solve[x^3 - 7 y^3 == 1, {x, y}, Integers]
Out[2]= {{x -> 1, y -> 0}, {x -> 2, y -> 1}}

Both are complete — \(x^3 - 2y^3 = 1\) has only \((1,0)\) and \((-1,-1)\), and \(x^3 - 7y^3 = 1\) adds \((2, 1)\) since \(8 - 7 = 1\). When the form represents \(m\) in no way at all, the empty set is again a proof:

In[3]:= Solve[x^3 - 2 y^3 == 5, {x, y}, Integers]
Out[3]= {}

3.2 Quartic and totally-complex forms

The engine is not limited to cubics. A quartic Thue form works the same way, and when the field is totally complex (no real embeddings) Mathilda uses an elementary rigorous bound \(|y| \le (|m| / \prod_i |\mathrm{Im}\,\theta_i|)^{1/n}\) in place of Baker's machinery — faster, and enough to close the solution set:

In[1]:= Length[Solve[x^4 + y^4 == 17, {x, y}, Integers]]
Out[1]= 8

In[2]:= Solve[x^4 - 2 y^4 == -1, {x, y}, Integers]
Out[2]= {{x -> -1, y -> -1}, {x -> -1, y -> 1}, {x -> 1, y -> -1}, {x -> 1, y -> 1}}

\(x^4 + y^4 = 17\) has the eight sign-and-swap variants of \((1, 2)\) and \((2, 1)\) — \(1 + 16 = 17\) — over the field \(\mathbb{Q}(\zeta_8)\); the Ljunggren-type \(x^4 - 2y^4 = -1\) has the four \((\pm 1, \pm 1)\). Thue equations Mathilda cannot yet reach (very large regulators, some non-monogenic fields, some \(|m| \ne 1\)) are declined, never guessed — a comparison against PARI/GP's thue() is on the performance page.

References

  1. A. Thue, Über Annäherungswerte algebraischer Zahlen, J. Reine Angew. Math. 135 (1909), 284–305 — the finiteness theorem.
  2. N. Tzanakis & B. M. M. de Weger, On the practical solution of the Thue equation, J. Number Theory 31 (1989), 99–132.
  3. A. Baker, Linear forms in the logarithms of algebraic numbers, Mathematika 13 (1966), 204–216.

Next: Exponential Diophantine equations — equations where the unknown sits in the exponent, decided by Mihăilescu's theorem and the Ramanujan–Nagell analysis.