On the Diophantine equation a^{3} + b^{3} + c^{3} + d^{3} = 0 by

RACHEL GAR-EL and

LEONID VASERSTEIN (University Park, PA)

Introduction. The equation

a^{3} + b^{3} + c^{3} + d^{3} = 0  (1)

has been studied by many mathematicians since Diophantus (see [B, p. 24]; [C]; [D, pp. 550–562]; [S]). Partial solutions in integers and complete solutions in rational numbers have been found.

A general solution of (1) found by Euler (see [H, pp. 290–291) involves the following polynomials s_{i} = s_{i}(x, y, z) :

s_{1} := 9x^{3} + 9x^{2}y + 3xy^{2} + 3y^{3} − 3x^{2}z + 6xyz − 3y^{2}z + 3xz^{2} + yz^{2} − z^{3}, s_{2} := 9x^{3} − 9x^{2}y + 3xy^{2} − 3y^{3} + 3x^{2}z + 6xyz + 3y^{2}z + 3xz^{2} − yz^{2} + z^{3}, s_{3} := −9x^{3} + 9x^{2}y − 3xy^{2} + 3y^{3} + 3x^{2}z + 6xyz + 3y^{2}z − 3xz^{2} + yz^{2} + z^{3}, s_{4} =: −9x^{3} − 9x^{2}y − 3xy^{2} − 3y^{3} − 3x^{2}z + 6xyz − 3y^{2}z − 3xz^{2} − yz^{2} − z^{3}. Using these polynomials, all rational solutions to (1) can be described in the following

way:

THEOREM 1. All rational solutions (a, b, c, d) of (1) up to nonzero rational factors are in 1-1 correspondence with all triples (x, y, z) up to nonzero rational factors according to the formulas (x, y, z) ↦ (a, b, c, d) = (s_{1}, s_{2}, s_{3}, s_{4}); and (a, b, c, d) ↦ (x, y, z) = (ac − bd, −a^{2} + ab − b^{2} + c^{2} − cd + d^{2}, a^{2} − ab + b^{2} − ac + 2bc + c^{2} + 2ad − bd − cd + d^{2}).

This theorem allows us to describe all integer solutions as follows:

COROLLARY 2. Up to a rational factor, every integral solution (a, b, c, d) of (1) is equal to (s_{1}, s_{2}, s_{3}, s_{4}) where s_{i} = s_{i}(x, y, z) are as above with integers x, y, z. Every integral primitive solution (a, b, c, d) of (1) can be written uniquely as (s_{1}, s_{2}, s_{3}, s_{4})/D with D = gcd(s_{1}, s_{2}, s_{3}, s_{4}) where s_{i} are as above with integral primitive (x, y, z).

Recall that an n-tuple (u_{1}, …, u_{n}) of integers is called primitive if gcd(u_{1}, …, u_{n}) = 1. Here gcd stands for the greatest common divisor which takes integral nonnegative values.

Corollary 2 does not give an explicit description of all integral primitive solutions to (1) since it is not clear from the definition D = gcd(s_{1}, s_{2}, s_{3}, s_{4}) what are possible values for D when (x, y, z) ranges over all integral primitive triples. So the complete solution of (1) in integers was pointed out as an open problem in [B, p. 10], [[C, p. 1251], [H, p. 290], [Ha, pp. 199–200], [R, v.3, p. 197], [S, pp. 121–122].

For comparison, let us consider a simpler Diophantine equation

a^{2} + b^{2} = c^{2}.  (2)

It is well-known that every primitive integral solution (a, b, c) of (2) can be written as ±(x^{2} − y^{2}, 2xy, x^{2} + y^{2})/d with a primitive pair (x, y) which is unique, up to sign, and

d = gcd(x^{2} − y^{2}, 2xy, x^{2} + y^{2}).

The explicit description of this number d is given as follows: d = mod(x, 2)+mod(y, 2). Here and further on, mod(x, 2) denotes the remainder on division of x by 2. So d = 2 when xy is odd, and d = 1 otherwise. Thus, the equation (2) is completely solved in integers.

In this paper we address the problem of finding an explicit description of the number D in Corollary 3. Our main result is:

THEOREM 3. Every integral primitive solution (a, b, c, d) of (1) can be written uniquely as (s_{1}, s_{2}, s_{3}, s_{4})/D with D =gcd(s_{1}, s_{2}, s_{3}, s_{4}) = d_{0}d_{2}d_{3} where polynomials s_{i} = s_{i}(x, y, z) are as above with integral primitive (x, y, z) and

d_{0} = gcd(x, 3y^{2} + z^{2})gcd(y, 3x^{2} + z^{2})gcd(z, 3x^{2} + y^{2}),

d_{2} = \{ \begin{matrix}4 & when mod(x, 2)+mod(y, 2)+mod(z, 2) = 2 and mod(xyz, 4) ≠ 0, \\ 2 & when mod(x, 2)+mod(y, 2)+mod(z, 2) = 2 and mod(xyz, 4) = 0, \\ 1 & otherwise, \end{matrix}

d_{3} = \{ \begin{matrix}3 & when mod(z, 3) = 0 and mod(xy, 3) ≠ 0, \\ 1 & otherwise. \end{matrix}

Our description of D is explicit enough to answer the question of what are its possible values.

COROLLARY 4. When (x, y, z) ranges over all primitive integral triples, the number D in Corollary 2 ranges over all numbers of the form t_{2}t_{3}t where t_{2} is 1 or 8, t_{3} is 1, 3, or 9, and t is any product of primes of the form 3k + 1.

Note that while the number d for the equation (2) is bounded, the number D for the equation (1) is not. While there are many polynomial solutions for (1) with integral coefficients besides (s_{1}, s_{2}, s_{3}, s_{4}), we believe that the set of primitive integral solutions of (1) cannot be covered by a finite set of polynomial families with integral coefficients.

1. Proof of Theorem 3. By Theorem 1, (s_{1}, s_{2}, s_{3}, s_{4}) = 0 if and only if (x, y, z) = 0. Assume now that (s_{1}, s_{2}, s_{3}, s_{4}) ≠ 0. Let D = gcd(s_{1}, s_{2}, s_{3}, s_{4}). Let p^{m} be a primary factor of D, i.e., p is a prime and p^{m} is the highest power of p dividing D.

We have to prove that p^{m} is equal to the highest power p^{n} of p dividing d_{0}d_{2}d_{3}. (By definition, D and all d_{i} are positive.)

Now we consider three cases:

Case 1 : p ≥ 5. First we prove that m ≤ n, i.e., p^{m}|d_{0}. If m = 0, there is nothing to prove, so let m ≥ 1.

Since p^{m}|s_{1}, we conclude: if p divides both x and y, then p|z^{3}, hence p divides z; if p divides both x and z, then p divides 3y^{3}; if p divides both z and y, then p divides 9x^{3}. Since (x, y, z) is primitive and p ≠ 3 , p divides at most one of the numbers x, y, z.

Since p^{m} divides s_{1} + s_{2} + s_{3} + s_{4} = 24xyz, we conclude that p^{m} divides x, y or z.

Subcase 1x : p ≥ 5 and p^{m} divides x. Then p^{m} divides
(s_{2} + s_{3})|_{x=0} = 2z(3y^{2} + z^{2})

hence p^{m}|3y^{2} + z^{2}. Therefore p^{m}|d_{0}.

Subcase 1y : p ≥ 5 and p^{m} divides y. Then p^{m} divides

(s_{2} + s_{3})|_{y=0} = 2z(3x^{2} + z^{2})

hence p^{m}|3x^{2} + z^{2}. Therefore p^{m}|d_{0}.

Subcase 1z : p ≥ 5 and p^{m} divides z. Then p^{m} divides

(s_{1} + s_{2})|_{z=0} = 6z(3x^{2} + y^{2})

hence p^{m}|3x^{2} + y^{2}. Therefore p^{m}|d_{0}.

Thus, p^{m}|d_{0} in all three subcases, i.e., m ≤ n in Case 1.

Since p^{n}|d_{0}, we conclude that either p^{n} divides both x and 3y^{2} + z^{2}, or it divides both y and 3z^{2} + x^{2} or else it divides both z and 3x^{2} + y^{2}. Therefore in all three cases it divides D, i.e., m ≥ n. Thus, m = n in Case 1.

Case 2 : p = 2. When only one or all three of (x, y, z) are odd, D is odd, so m = 0. In this case, also all three d_{0}, d_{2}, d_{3} are odd so n = 0 = m.

Assume now that exactly 2 of the numbers (x, y, z) are odd, i.e.,

mod(x, 2) + mod(y, 2) + mod(z, 2) = 2.

Then m = 3, and n = 3 as well.

Case 3 : p = 3. When mod(z, 3) ≠ 0, m = n = 0. When mod(z, 3) = 0 and mod(y,

3) ≠ 0, then m = 1, and also n = 1. When mod(z, 3) = 0 = mod(y, 3), then m = 2, and also n = 2.

2. Proof of Corollary 4. In the proof of Theorem 3 in Cases 2 and 3, we saw that the power of 2 dividing D can be 1 or 8, and the power of three dividing D can be 1, 3 or 9. Now let p be a prime ≥ 5 dividing D. Modulo such a prime, −3 is a square, therefore the multiplicative group modulo p contains an element of order 3, so p − 1 is divisible by 3, i.e., p = 3k + 1. Therefore D has the form described in Corollary 6.

Now it remains to be proved that every number D of this form can actually occur. We write D = 2^{m}3^{n}d′ where m = 0 or 3, n = 0, 1, or 2, and d′ is a product of primes of the form 3k + 1.

We take x = d′. We set

y = \{ \begin{matrix}1 & when n ≤ 1, \\ 3 & otherwise. \end{matrix}

Note that −3 is a square modulo every prime of the form 3k + 1, and so is −3y^{2}. By the Chinese remainder theorem, we can choose z such that mod(z^{2} + 3y^{2}, x) = 0, i.e., gcd(x, 3y^{2}+z^{2}) = x = d′. In addition, we can impose on z any congruences modulo 4 and 3. We require z to be odd when m = 0 and mod(z, 4) = 0 otherwise. We require mod(z, 3) = mod(D, 3). Still we can replace z by z+12ux with any integer u keeping all these conditions intact. Since gcd(3x^{2} + y^{2}, 12x) divides 12, we can arrange that gcd(z, 3x^{2} + y^{2})|12.

Then:

gcd(z, 3x^{2} + y^{2}) = \{ \begin{matrix}1 & when n ≤ 1 and m = 0, \\ 3 & when n = 2 and m = 0, \\ 4 & when n ≤ 1 and m ≥ 1, \\ 12 & otherwise, \end{matrix}

gcd(x, 3y^{2} + z^{2}) = d′,

and

gcd(y, 3x^{2} + z^{2}) = \{ \begin{matrix}1 & when n ≤ 1, \\ 3 & otherwise. \end{matrix}
Therefore d_{0} = 2^{m′}3^{n′}d′ where

m′ = \{ \begin{matrix}0 & when m = 0, \\ 2 & when m = 3, \end{matrix}

and

n′ = \{ \begin{matrix}0 & when n ≤ 1, \\ 2 & when n = 2, \end{matrix}

hence D = d_{2}d_{3}d′.

References

[B] B. C. Berndt, Y.-S. Choi, and S.-Y. Kang, The problems submitted by Ramanujan to the Journal of the Indian Mathematical Society, in: Proceedings of conference, “Continued Fractions: From Analytic Number Theory to Constructive Approximation,” held at the University of Missouri in May, 1998; B. C. Berndt and F. Gesztesy, editors, AMS Cont.Math. 236 (1999), 15–56.

[C] A. Choudhry, On equal sums of cubes, Rocky Mountain Journal of Math. 28:4 (1998), 1251–1257.

[D] L.E. Dickson, History of the theory of numbers, New York, Chelsea Pub. Co., (1966), v. 2. Diophantine analysis.

[H] L.-K. Hua, Introduction to number theory, Berlin; New York, Springer-Verlag, 1982.

[Ha] G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, 5th ed. Oxford, Clarendon Press; New York, Oxford University Press, 1979.

[R] Ramanujan’s notebooks, edited by B C. Berndt, New York, Springer-Verlag, 1985– 1998, 5 v.

[S] C. Sándor, On the equation a^{3} + b^{3} + c^{3} = d^{3}, Period. Math. Hungar. 33:2 (1996), 121–134.

Department of Mathematics Penn State University University Park, Pennsylvania 16802 U.S.A. E-mail: vstein@math.psu.edu

gar-el@math.psu.edu