Proceedings of the Steklov Institute of Mathematics, Vol. 218, 1997, pp. 94–103. Translated from Trudy Matematicheskogo Instituta imeni V. A. Steklova, Vol. 218, 1997, pp. 99–108. Original Russian Text Copyright \begin{matrix}c \\ ◯ \end{matrix} 1997 by Garaev. English Translation Copyright \begin{matrix}c \\ ◯ \end{matrix} by ���� �����/Interperiodica Publishing (Russia).

Third-Degree Diophantine Equations

M. Z. Garaev

Received February, 1997

1. INTRODUCTION

This work is devoted to the investigation of the solvability of the Diophantine equation

x^{3} + y^{3} + z^{3} = nxyz,  (1)

where n ∉ {0, 3}, n is an integer. The first results were obtained as long ago as 1856 by Sylvester (see [1]). He proved that for n = −6 (1) had no solutions in nonzero integers.

The next results were obtained almost eighty years later. In 1933–34 Hurwitz and Mordell showed that for n = −1 and for n = 5 (1) had the only solutions (1, 1, −1) and (1, 1, 2), respectively, with an accuracy to within the permutation and to the common factor (see [2, 3]).

Developing Mordell’s method, Cassels showed in 1960 that for n = 1 (1) had no solutions in nonzero integers (see [4, 5]).

Note that for the integer n (1) is solvable in nonzero positive integers x, y, z if and only if the equation

\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = n  (2)

is solvable in positive integers.

Considering (2) for n ≥ 4, Serpinsky notes in [6] that he does not know whether this equation has a solution in positive integers for n = 4. It follows from Theorem 1, Section 2 of this work that (2) has no solutions in the positive integers x, y, z not only for n = 4 but also for any values of n which are multiples of 4 as well as for many odd values of n > 4.

In Section 3 we consider various Diophantine equations which are connected, to a certain extent, with (1).

2. ON THE EQUATION x^{3} + y^{3} + z^{3} = nxyz

Theorem 1. Let n ∈ {4k, 8k − 1, 2^{2m+1}(2k − 1) + 3}, where k and m run through all positive integers. Then (1) has no solutions in the positive integers x, y, z.

Corollary. For the values of n indicated in Theorem 1 (2) has no solutions in the positive integers x, y, z.

In order to prove the theorem, we shall need some auxiliary statements which we shall formulate as lemmas. Lemma 1 belongs to Sylvester (see [1]).

Lemma 1. Let A, B, C, D, α, β, γ be arbitrary real numbers for which

Aα^{3} + Bβ^{3} + Cγ^{3} = Dαβγ.

Then

ABCf^{3} + g^{3} + h^{3} = Dfgh,

where

h = A^{2}Bα^{6}β^{3} + B^{2}Cβ^{6}γ^{3} + C^{2}Aγ^{6}α^{3} − 3ABCα^{3}β^{3}γ^{3},
g = AB^{2}α^{3}β^{6} + BC^{2}β^{3}γ^{6} + CA^{2}γ^{3}α^{6} − 3ABCα^{3}β^{3}γ^{3},
f = αβγ(A^{2}α^{6} + B^{2}β^{6} + C^{2}γ^{6} − ABα^{3}β^{3} − ACα^{3}γ^{3} − BCβ^{3}γ^{3}).

Consider the functions

f(x, y, z) = xyz(x^{6} + y^{6} + z^{6} − x^{3}y^{3} − y^{3}z^{3} − z^{3}x^{3}),
g(x, y, z) = x^{3}y^{6} + y^{3}z^{6} + z^{3}x^{6} − 3x^{3}y^{3}z^{3},
h(x, y, z) = x^{6}y^{3} + y^{6}z^{3} + z^{6}x^{3} − 3x^{3}y^{3}z^{3}.

In this notation, Lemma 1 means that if (1) is valid, then

f^{3}(x, y, z) + g^{3}(x, y, z) + h^{3}(x, y, z) = nf(x, y, z)g(x, y, z)h(x, y, z).

Lemma 2. Suppose that n ≠ 3, n is an integer, and (x_{1}, y_{1}, z_{1}) is the solution of (1). Then

(a) if x_{1}, y_{1}, z_{1} are positive integers, then f(x_{1}, y_{1}, z_{1}), g(x_{1}, y_{1}, z_{1}), h(x_{1}, y_{1}, z_{1}) are also positive integers,

(b) if x_{1}, y_{1}, z_{1} are nonzero integers, then f(x_{1}, y_{1}, z_{1}), g(x_{1}, y_{1}, z_{1}), h(x_{1}, y_{1}, z_{1}) are also nonzero integers.

Proof. Item (a) immediatly follows from the relations

u^{3} + v^{3} + w^{3} − 3uvw = \frac{1}{2}(u + v + w)((u − v)^{2} + (v − w)^{2} + (w − u)^{2}),

u^{2} + v^{2} + w^{2} − uv − vw − wu = \frac{1}{2}((u − v)^{2} + (v − w)^{2} + (w − u)^{2}).

Item (b) follows from the fact that if (1) is valid, where x, y, z, n are integers, then (x, y) = (y, z) = (z, x) = (x, y, z). Therefore, if xy^{2} + yz^{2} + zx^{2} = 0, then |x| = |y| = |z|, x + y + z = 0, but this is impossible for xyz ≠ 0.

Let n ≠ 3, (x_{1}, y_{1}, z_{1}) be a triple of nonzero integers satisfying (1), and GCD(x_{1}, y_{1}, z_{1}) = 1 (hence, (x_{1}, y_{1}) = (y_{1}, z_{1}) = (z_{1}, x_{1}) = 1).

We define the sequence of the triples (x_{i}, y_{i}, z_{i}) as follows:

x_{k+1} = \frac{f_{k}}{(f_{k}, g_{k}, h_{k})},  y_{k+1} = \frac{g_{k}}{(f_{k}, g_{k}, h_{k})},  z_{k+1} = \frac{h_{k}}{(f_{k}, g_{k}, h_{k})},

where

f_{k} = f(x_{k}, y_{k}, z_{k}),  g_{k} = g(x_{k}, y_{k}, z_{k}),  h_{k} = f(x_{k}, y_{k}, z_{k}),

and (f_{k}, g_{k}, h_{k}) := GCD(f_{k}, g_{k}, h_{k}).

It follows from what was said above that the following lemma is valid.

96 GARAEV

Lemma 3. For any positive integer i the triple (x_{i}, y_{i}, z_{i}) is the solution of (1) and (x_{i}, y_{i}) = (y_{i}, z_{i}) = (z_{i}, x_{i}) = 1.

Lemma 4. For any positive integer i the congruence

x_{i+1} ≡ 0 (mod(x_{i}y_{i}z_{i}))

holds true.

Proof. By definition,

(f_{i}, g_{i}, h_{i})x_{i+1} = f_{i} = x_{i}y_{i}z_{i}(x_{i}^{6} + y_{i}^{6} + z_{i}^{6} − x_{i}^{3}y_{i}^{3} − y_{i}^{3}z_{i}^{3} − z_{i}^{3}x_{i}^{3}),

and therefore

(f_{i}, g_{i}, h_{i})x_{i+1} ≡ 0 (mod(x_{i}y_{i}z_{i}))

and, all the more so,

g_{i}x_{i+1} ≡ 0 (mod(x_{i}y_{i}z_{i})).

If we assume that x_{i+1} ≢ 0 (mod(x_{i}y_{i}z_{i})), then there exists a prime number p for which x_{i}y_{i}z_{i} ≡ 0 (mod p), g_{i} ≡ 0 (mod p). Suppose, for instance, that x_{i} ≡ 0 (mod p). Then it follows from

x_{i}^{3}y_{i}^{6} + y_{i}^{3}z_{i}^{6} + z_{i}^{3}x_{i}^{6} − 3x_{i}^{3}y_{i}^{3}z_{i}^{3} = g_{i} ≡ 0 (mod p)

that y_{i}^{3}z_{i}^{6} ≡ 0 (mod p), i.e., either y_{i} or z_{i} is divisible by p, and this, together with x_{i} ≡ 0 (mod p), contradicts Lemma 3.

A simillar contradiction can be found in the case where y_{i} ≡ 0 (mod p) or z_{i} ≡ 0 (mod p). Lemma 5. If i ≠ j, then

(y_{i}z_{i}, y_{j}z_{j}) = 1.  (3)

Proof. Indeed, suppose, for definiteness, that i < j. Then Lemma 4 yields x_{j} ≡ 0 (mod(y_{i}z_{i})). According to Lemma 3, (x_{j}, y_{j}z_{j}) = 1. Therefore relation (3) holds true. Lemma 6. For any positive integer N there exists a number i such that (y_{i}z_{i}, N) = 1. Proof. Indeed, let N = p_{1}^{k_{1}} ⋯ p_{s}^{k_{s}} be a prime factorization of the number N. The statement of

the lemma follows from the fact that the terms of the sequence of s + 1 numbers y_{1}z_{1}, …, y_{s+1}z_{s+1} are, according to Lemma 5, pairwise prime numbers.

Thus the following fundamental lemma is valid. Fundamental lemma. Let n be an integer and let (1) be solvable in nonzero positive integers.

Then, for any positive integer N, there exists a triple (x_{N}, y_{N}, z_{N}) of nonzero positive integers which is the solution of (1) and which satisfies the conditions

(x_{N}, y_{N}) = (y_{N}, z_{N}) = (z_{N}, x_{N}) = (y_{N}z_{N}, N) = 1.

Lemma 7. Let n be an integer, n ≠ 3, and let x, y, z be nonzero pairwise prime integers satisfying relation (1). Suppose, furthermore, that d = (x + y, z), i.e.,

x + y = da, z = db, (a, b) = 1.  (4)

Then there exists an integer α such that

n^{3} − 27 ≡ 0 (mod α),  (5)
(α, a) = (α, b) = 1,  (6)
4 ≡ 0 (mod(α, nb − a)),  (7)
α(4z^{2} + n(x − y)^{2}) = a(n^{2}ab − na^{2} + 12b^{2}),  (8)
α(z^{2} + n(x − y)^{2}) = (nb − a)(na^{2} − 3b^{2}).  (9)

When x, y, z are positive integers, the numbers α, a, b, nb − a are also positive integers.

Proof. According to the hypothesis,

(x + y)^{3} + z^{3} = xy(3(x + y) + nz), x + y = da, z = db, (a, b) = 1, d = (x + y, z),

whence it follows that

d^{2}(a^{3} + b^{3}) = (3a + nb)xy.

Since (x, y) = (x + y, xy) = 1, we note that (d^{2}, xy) = 1. Therefore there exists an integer α such that

\{ \begin{matrix}3a + nb = αd^{2} \\ a^{3} + b^{3} = αxy. \end{matrix}  (10)

We write the second relation of system (10) in the form

4a^{3} + 4b^{3} = 4αxy = α(x + y)^{2} − α(x − y)^{2} = αd^{2}a^{2} − α(x − y)^{2} = (3a + nb)a^{2} − α(x − y)^{2}.

Consequently,

\{ \begin{matrix}αd^{2} = 2a + nb \\ α(x − y)^{2} = na^{2}b − a^{3} − 4b^{3}, \end{matrix}

whence, after elementary transformations, we get (8), (9).

We can see from system (10) that (6) is satisfied for α and

(n^{3} − 27)b^{3} ≡ 0 (mod α),

i.e., (5) is satisfied as well.

Next it follows, also from (10), that 4a is divisible by (α, nb − a). Together with (α, a) = 1 this means that relation (7) holds true.

If x, y, z are positive integers, then d, a, b are positive integers, and therefore α is also a positive integer (n ≥ 4 from (10) ) In addition, in this case we have

d(nb − a) = nz − (x + y) = \frac{x^{3} + y^{3} + z^{3}}{xy} − (x + y) = \frac{(x + y)(x − y)^{2} + z^{3}}{xy} > 0,

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

i.e., nb − a > 0. We proved the lemma.

Proof of Theorem 1. We shall carry out the proof by contradiction. Let (1) be solvable in positive integers x, y, z for a certain n from the hypothesis of the theorem. Taking N = 2n in the fundamental lemma, we find that (1) is solvable in positive integers which we shall also denote by x, y, z and which satisfy the conditions

(x, y) = (y, z) = (z, x) = 1, (yz, 2n) = 1.

Using Lemma 7, we have (4), and there exists a positive integer α for which (5)–(7) are satisfied and

α(4z^{2} + n(x − y)^{2}) ≡ 0 (mod a),  (11)
α(z^{2} + n(x − y)^{2}) ≡ 0 (mod(nb − a)),  (12)
(yz, 2n) = 1.  (13)

It is easy to note that (n, a) = (n, nb − a) = 1. Indeed, if p is a prime number or unity and is such that n ≡ 0 (mod p), a ≡ 0 (mod p), then

x + y = da ≡ 0 (mod p),  (14)

and z ≡ 0 (mod p) according to (1) . From (13) we find that p = 1. This means that (n, nb − a) = (n, a) = 1.

We divide the possible values of n into three variants, namely,

(1) n = 4k,

(2) n = 8k − 1,

(3) n = 2^{2m+1}(2k − 1) + 3.

(1) Since (α, a) = 1, it follows from (11) that

4z^{2} + n(x − y)^{2} ≡ 0 (mod a).  (15)

It follows from (5) that α is an odd number, and therefore conditions (7) and (12) mean that

z^{2} + n(x − y)^{2} ≡ 0 (mod(nb − a)),  (16)

It follows from (13) that y, z are odd numbers and (1) implies the evenness of the number x (n = 4k). All this means that the numbers a and b are odd.

Let us show that

(x − y, a) = (x − y, nb − a) = 1.  (17)

Indeed, if p is a prime number or unity and x−y ≡ 0 (mod p), a ≡ 0 (mod p), then relation (14) is satisfied, whence it follows that 2x ≡ 0 (mod p), 2y ≡ 0 (mod p). Since (x, y) = 1, we get either p = 2 or p = 1. The case p = 2 cannot hold since a is an odd number. Only the case p = 1 remains. This means that (x − y, a) = 1.

Now if x − y ≡ 0 (mod p), nb − a ≡ 0 (mod p), then we find from (16) that z ≡ 0 (mod p), x^{3} + y^{3} = nxyz − z^{3} ≡ 0 (mod p). Therefore 2x^{3} ≡ 0 (mod p), 2y^{3} ≡ 0 (mod p), i.e., either p = 2 or p = 1. Since z is odd, only p = 1 remains.

We have thus proved (17). Taking this into account in congruences (15) and (16), we find that there exist positive integers z_{1} and z_{2} for which

z_{1}^{2} ≡ −n (mod a), z_{2}^{2} ≡ −n (mod(nb − a)).

Next, n = 4k, a is odd, nb − a is odd, (n, a) = (n, nb − a) = 1. Therefore

(\frac{−n}{a}) = (\frac{−n}{nb − a}) = 1,

where (\frac{p}{q}) is Jacobi’s symbol.

Let us write n as n = 2^{m}n_{0}, m ≥ 2, where m is a positive integer and n_{0} is an odd positive integer,

1 = (\frac{−n}{a}) = (−1)^{(a−1)/2}(\frac{2^{m}n_{0}}{a} ) = (−1)^{(a−1)/2}(−1)^{(a^{2}−1)m/8}(\frac{n_{0}}{a} ).

Therefore

(\frac{a}{n_{0}} ) = (−1)^{(a−1)/2}(−1)^{(a^{2}−1)m/8}(\frac{n_{0}}{a} )(\frac{a}{n_{0}} ) = (−1)^{(a−1)/2+(a^{2}−1)m/8+(a−1)(n_{0}−1)/2}.

Similarly, we find from (\frac{−n}{nb−a}) = 1 that

(\frac{nb − a}{n_{0}}) = (−1)^{(nb−a−1)/2+((nb−a)^{2}−1)m/8+(nb−a−1)(n_{0}−1)/2}
= (−1)^{(−a−1)/2+(−2^{m+1}n_{0}ab+a^{2}−1)m/8−(a+1)(n_{0}−1)/2}.

Taking into account that 2^{m+1}m/8 is an odd number for m ≥ 2, n ≡ 0 (mod n_{0}), we have

(−1)^{(n_{0}−1)/2} = (\frac{a}{n_{0}} )(\frac{nb − a}{n_{0}})
= (−1)^{(a−1)/2+(a^{2}−1)m/8+(a−1)(n_{0}−1)/2−(a+1)/2+(a^{2}−1)m/8−(a+1)(n_{0}−1)/2}
= (−1)^{1+(n_{0}−1)/2} = −(−1)^{(n_{0}−1)/2},

i.e., (−1)^{(n_{0}−1)/2} = −(−1)^{(n_{0}−1)/2}.

We have obtained a contradiction which proves the theorem for case (1).

(2) Let a = 2^{s_{1}}a_{1}, nb − a = 2^{s_{2}}b_{1}, where s_{1} ≥ 0, s_{2} ≥ 0 are integers and a_{1}, b_{1} are odd positive integers. From (6), (7), (11), (12) we obtain

4z^{2} + n(x − y)^{2} ≡ 0 (mod a_{1}), z^{2} + n(x − y)^{2} ≡ 0 (mod b_{1}).

Just as in case (1), it is easy to show that (x − y, a_{1}) = (x − y, b_{1}) = 1. Taking into account that (n, a_{1}) = (n, b_{1}) = 1, we again have

(\frac{−n}{a_{1}}) = (\frac{−n}{b_{1}}) = 1,

(\frac{a_{1}}{n} ) = (\frac{−n}{a_{1}})(\frac{a_{1}}{n} ) = (−1)^{(a_{1}−1)/2}(−1)^{((a_{1}−1)/2)((n−1)/2)} = (−1)^{(a_{1}−1)/2}(−1)^{(a_{1}−1)/2} = 1,

i.e., (\frac{a_{1}}{n} ) = 1.

Similarly, (\frac{b_{1}}{n} ) = 1. Since n = 8k − 1, it follows that (\frac{2}{n}) = (−1)^{(n^{2}−1)/8} = 1. Therefore

(\frac{a}{n}) = (\frac{2^{s_{1}}a_{1}}{n} ) = 1, (\frac{nb − a}{n}) = (\frac{2^{s_{2}}b_{1}}{n} ) = 1,

whence it follows that

1 = (\frac{a}{n})(\frac{nb − a}{n}) = (−1)^{(n−1)/2}(\frac{a}{n})(\frac{a}{n}) = −1.

The contradiction obtained proves the theorem for case (2).

(3) We shall first prove that the number x is even. Let us assume that this is not the case. Then, since (yz, 2n) = 1, all numbers x, y, z are odd.

We rewrite (1) in the form

(x + y + z)\frac{(2x − y − z)^{2} + 3(y − z)^{2}}{4} = 2^{2m+1}(2k − 1)xyz.

Since the numbers x + y + z and xyz are odd, we have

u^{2} + 3v^{2} ≡ 0 (mod 2^{2m+3}), u^{2} + 3v^{2} ≡ 0 (mod 2^{2m+4}),

where u = 2x − y − z, v = y − z. It is easy to see that this is impossible for any integers u, v and any positive integers m.

Thus x is an even number and yz is an odd number. Next, it follows from the relation

(x + y + z)(x^{2} + y^{2} + z^{2} − xy − yz − zx) = 2^{2m+1}(2k − 1)xyz

that x + y + z ≡ 0 (mod 16).

Therefore, since z and d are odd, we find from the relation

d(nb − a) = nz − (x + y) = (2^{2m+1}(2k − 1) + 4)z − (x + y + z)

that the number nb − a, which is divisible by 4, is not divisible by 8.

Thus a is an odd number and nb − a = 4b_{1}, where b_{1} is also an odd number. Just as in cases (1) and (2), we have

(α, a) = (α, b_{1}) = 1,  (x − y, a) = (x − y, b_{1}) = 1,  (n, a) = (n, b_{1}) = 1

and

(\frac{−n}{a}) = (\frac{−n}{b_{1}}) = 1,

(\frac{a}{n}) = (\frac{−n}{a})(\frac{a}{n}) = (−1)^{(a−1)/2}(−1)^{((a−1)/2)((n−1)/2)} = (−1)^{(a−1)/2}(−1)^{(a−1)/2} = 1,

i.e., (\frac{a}{n}) = 1. Similarly, (\frac{b_{1}}{n} ) = 1.

Therefore

−1 = (−1)^{(n−1)/2} = (\frac{−1}{n}) = (\frac{−a}{n}) = (\frac{nb − a}{n}) = (\frac{4b_{1}}{n} ) = (\frac{b_{1}}{n} ) = 1.

The contradiction obtained proves the theorem in case (3). This completes the proof of Theorem 1.

The corollary can be proved either with the use of Lemma 1 or in any other way.

3. ON THE DIOPHANTINE EQUATION x^{3} + y^{3} + 1 = nxy

Lemma 8. If the relation

x_{0}^{3} + y_{0}^{3} + 1 = n_{0}x_{0}y_{0}

is satisfied for the positive integers x_{0}, y_{0}, n_{0}, then

x_{1}^{3} + y_{1}^{3} + 1 = n_{1}x_{1}y_{1},  (18)

where

x_{1} = n_{0}x_{0} − y_{0}^{2},  y_{1} = x_{0},  n_{1} = n_{0}^{2}x_{0} − n_{0}y_{0}^{2} − y_{0}x_{0}^{2},

where x_{1}, y_{1}, z_{1} are positive integers. In this case, if x_{0} ≥ y_{0}, then x_{1} > y_{1}, n_{1} > n_{0}.

This lemma can be proved by means of simple calculations.

According to Theorem 1, if n ∈ {4k, 8k − 1, 2^{2m+1}(2k − 1) + 3}, where k and m run through all positive integers, then the equation

x^{3} + y^{3} + 1 = nxy  (19)

has no solutions in positive integers.

Theorem 2. For any i ∈ {1, 2, 3, 5, 6} there exists an infinite number of positive integers n of the form 8k + i for which (19) has a solution in the positive integers x, y.

Proof. Let (x, y, n) = (x_{0}, y_{0}, n_{0}) be triples of positive integers satisfying (19), x_{0} ≥ y_{0}. We shall define the sequence (x_{j}, y_{j}, n_{j}) of triples of positive integers in accordance with Lemma 8:

x_{j} = n_{j−1}x_{j−1} − y_{j−1}^{2}, y_{j} = x_{j−1}, n_{j} = n_{j−1}^{2}x_{j−1} − n_{j−1}y_{j−1}^{2} − y_{j−1}x_{j−1}^{2}, j = 1, 2, ….

According to Lemma 8,

x_{j}^{3} + y_{j}^{3} + 1 = n_{j}x_{j}y_{j}, j = 1, 2, …, n_{0} < n_{1} < n_{2} < · · · .

We take x_{0} = 3, y_{0} = 2, n_{0} = 6 and write the remainders of the division of the sequence of triples (x_{j}, y_{j}, n_{j}) by 8 in a row. Obviously, the remainders can be calculated from the same formulas which define the sequence (x_{j}, y_{j}, n_{j})

(3, 2, 6) → (6, 3, 2) → (3, 6, 2) → (2, 3, 6) → (3, 2, 6) → · · · .

We see that the remainders are repeated, and we have

n_{4j} ≡ n_{4j−1} ≡ 6 (mod 8), n_{4j+1} ≡ n_{4j+2} ≡ 2 (mod 8)

for any positive integer j.

And now we take x_{0} = 1, y_{0} = 1, n_{0} = 3 and carry out the same procedure. We get

(1, 1, 3) → (2, 1, 5) → (1, 2, 1) → (5, 1, 3) → (6, 5, 1) → (5, 6, 1)
→ (1, 5, 3) → (2, 1, 1) → (1, 2, 5) → (1, 1, 3) → · · · .

In this case we have

n_{9j} ≡ n_{9j+3} ≡ n_{9j+6} ≡ 3 (mod 8), n_{9j+1} ≡ n_{9j+8} ≡ 5 (mod 8),
n_{9j+2} ≡ n_{9j+4} ≡ n_{9j+5} ≡ n_{9j+7} ≡ 1 (mod 8).

We have proved the theorem.

Theorem 3. The equation

x^{3} + 64y^{3} + 1 = 4xyz  (20)

has no solutions in positive integers x, y, z.

Proof. We shall carry out the proof by contradiction. Let the positive integers x, y, z be such that (20) is satisfied. According to Lemma 8, relation (18) holds true, where

x_{1} = zx − 16y^{2},  y_{1} = x,  n_{1} = z^{2}x − 16zy^{2} − 4yx^{2},

x_{1}, y_{1}, n_{1} are positive integers.

If z is an even number, then n_{1} is divisible by 4, and in that case relation (18) contradicts Theorem 1. Therefore z is an odd number. Then z^{2} ≡ 1 (mod 8). This means, by virtue of the oddness of x, that n_{1} ≡ x − 4y (mod 8).

It follows from (20) that x^{3} + 1 ≡ 4y (mod 8). Hence,

n_{1} ≡ x − 4y ≡ x − x^{3} − 1 ≡ −1 (mod 8),

i.e., n_{1} ≡ −1 (mod 8).

The last congruence, together with relation (18), contradicts Theorem 1.

Theorem 4. The equation

t^{2} = 4xyzuv − xu^{2} − yv^{2}  (21)

has no solutions in positive integers t, x, y, z, u, v.

Proof. We carry out the proof by contradiction. Let the positive integers t, x, y, z, u, v be such that (21) is satisfied.

Obviously, we can assume, without loss of generality, that (u, v) = 1. Then one of the numbers u, v is odd.

Let u be an odd number. We denote by d the greatest common divisor of the numbers u and y, i.e., let

u = du_{1},  y = dy_{1},  (u_{1}, y_{1}) = 1,  t^{2} = d(4xy_{1}zdu_{1}v − xdu_{1}^{2} − y_{1}v^{2}).

We set a = (d, y_{1}), i.e.,

d = ad_{1},  y_{1} = ay_{2},  (d_{1}, y_{2}) = 1.

We have

t^{2} = a^{2}d_{1}(4xy_{2}zad_{1}u_{1}v − xd_{1}u_{1}^{2} − y_{2}v^{2}),

\begin{aligned}(u, v) = 1 & ⟹ (d_{1}, v) = 1, (d_{1}, y_{2}) = 1, (d_{1}, 4xy_{2}zad_{1}u_{1}v − xd_{1}u_{1}^{2} − y_{2}v^{2}) = 1 \\ & ⟹ d_{1} = d_{2}^{2}, 4xy_{2}zad_{2}^{2}u_{1}v − xd_{2}^{2}u_{1}^{2} − y_{2}v^{2} = t_{1}^{2}, \end{aligned}

where d_{2}, t_{1} are certain positive integers.

We denote d_{2}u_{1} = u_{2}, zad_{2} = z_{1} and obtain

t_{2}^{2} = 4xy_{2}z_{1}u_{2}v − xu_{2}^{2} − y_{2}v^{2}.

In addition to the fact that u_{2} is an odd number, (u_{2}, v) = 1, we have obtained the equality (u_{2}, y_{2}) = 1.

THIRD-DEGREE DIOPHANTINE EQUATIONS 103

Next,

t_{2}^{2} ≡ −y_{2}v^{2} (mod(u_{2}(4y_{2}z_{1}v − u_{2}))),

whence it follows that

(\frac{−y_{2}}{u_{2}} ) = (\frac{−y_{2}}{4y_{2}z_{1}v − u_{2}} ) = 1.

Reasoning as in case (1) of Section 2, we arrive at a contradiction. Corollary. Let the integers a, b, c, d be such that

ab ≡ c ≡ d ≡ 1 (mod 4).

Then the equation

(a^{2}x)^{3} + (b^{2}y)^{3} + 1 = ab(cx + dy)xy  (22)

has no solutions in positive integers.

Proof. Indeed, if this is not the fact, i.e., if the positive integers x, y satisfy (22), then

(a^{2}x)^{3} + (b^{2}y)^{3} + 1 ≡ 0 (mod(4abxy)),

Hence, denoting t = (a^{2}x + b^{2}y)^{2}, we obtain

t^{2} ≡ −xa^{2} − yb^{2} (mod(4abxy)),

and this contradicts Theorem 4.

In conclusion, I want to express my deep gratitide to Professor V. N. Chubarikov for his scientific supervision.

ACKNOWLEDGMENTS

The article was written during my work on probation at the Mechanical and Mathematical Faculty of the Moscow State University under the guidance of Professor V. N. Chubarikov.

REFERENCES

1. Dickson, L. E., History of the Theory of Numbers, vol. 2, New York, 1934.

2. Hurwitz, A., Zahlentheorie, Algebra und Geometrie, in: Mathematische Werke, Basel, 1933, vol. 2.

3. Mordell, L. J., The Diophantine Equation x^{3} + y^{3} + z^{3} + kxyz = 0, in: Colloque sur la theorie des nombers, Bruxelles, 1955, pp. 67–76.

4. Cassels, J. W. S., On a Diophantine Equation, Acta Arith., 1960, vol. 6, pp. 47–51.

5. Serpinskii, V., 250 zadach po elementarnoi teorii chisel (250 Problems in Elementary Number Theory), Moscow:

Prosveshchenie, 1968.

Translated by I. Aleksandrova