The Diophantine equation X(exp 3) = 1 + 9v over quadratic fields

The Diophantine equation X

3

= 1 + 9v

over quadratic fields

Takaaki KAGAWA

LetK be a quadratic field and O_K,O_K× the ring of integers, the group of units, respectively. In this paper, we solve the Diophantine equation X3 = 1 + 9v in X ∈ O_K,v ∈ O_K×.

Keywords: Diophantine equation, Quadratic field

E-mail: [email protected] (T. Kagawa)

Department of Mathematical Sciences College of Science and Engineering

Ritsumeikan University Kusatsu, Shiga, 525–8577, Japan

立 命 館 大 学 理 工 学 研 究 所 紀 要 第77号 2018年

Memoirs of the Institute of Science and Engineering, Ritsumeikan University, Kusatsu, Shiga, Japan. No. 77, 2018

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1 Results

In the entrance examination for Ritsumeikan university [2], the following question was posed: Solve the Diophantine equation

x3_{+ 3xy + (1 − y}3_{) = 81 (1)}

inx, y ∈ Z. In this paper, we first solve this question:

Proposition. The only solutions (x, y) ∈ Z × Z of (1) are (x, y) = (−2, −4), (4, 2), (4, −4). We give an application of Proposition to the Diophantine equation

X3 _{= 1 + 9v (2)}

over quadratic fields.

Theorem. Let K be a quadratic field. Let OK be the ring of integers and OK× the group of

units.

(a) When K = Q(√−3), Q(√6), the only solution of (2) in (X, v) ∈ O_K× O×_K is (X, v) =

(−2, −1).

(b) When K = Q(√−3), the only solutions of (2) in (X, v) ∈ O_K × O_K× are (X, v) =

(−2, −1), (1 ±√−3, −1).

(c) When K = Q(√6), the only solutions of (2) in (X, v) ∈ O_K × O×_K are (X, v) =

(−2, −1), (−2 ±√6, −5 ± 2√6).

2 Proofs

Proof of Proposition. Factoring the left-hand side of (1) results in (x − y + 1)(x2+y2+ 1 +

xy + y − x) = 81, whence



x − y + 1 = 3a_ε,

x2_+y2_{+ 1 +xy + y − x = 3}4−a_ε,

where 0≤ a ≤ 4, ε = ±1. Eliminating x results in

3y2+ (3a+1ε − 3)y + (32a− 3a+1ε − 34−aε + 3) = 0.

The left-hand side of this equation is not divisible by 3 when a = 0 or a = 4. Thus we may suppose that a = 1, 2 or 3, and thus

y2_{+ (3}a_{ε − 1)y + (3}2a−1_{− 3}a_{ε − 3}3−a_{ε + 1) = 0. (3)}

If (3) has a solution y ∈ Z, then the discriminant

Dε(a) = −32a−1+ 2× 3aε + 4 × 33−aε − 3

Takaaki KAGAWA

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of (3) must be a square in Z. The only square values of D_ε(a) are D₁(1) = 36, D₁(2) = 0. We therefore have (x, y) = (−2, −4), (4, 2), (4, −4).

Proof of Theorem. Taking norm of (2), we obtain

NK/Q(X)3+ 3NK/Q(X) TrK/Q(X) + (1 − TrK/Q(X)3) = 81NK/Q(v). (4)

Reducing modulo 4 shows that the left-hand side of (4) is not congruent to 3 modulo 4. Thus

NK/Q(v) = 1 and thus Proposition yields (NK/Q(X), TrK/Q(X)) = (−2, −4), (4, 2), (4, −4),

that is, (X, v) = (−2 ±√6, −5 ± 2√6), (1 ±√−3, −1), (−2, −1).

Remark. In [1], the Diophantine equation X3 _{= 1 + 27v is solved in the same way.}

References

[1] T. Kagawa, Nonexistence of elliptic curves having everywhere good reduction and cubic discriminant, Proc. Japan Acad. 76, Ser. A (2000), 141–142.

[2] Entrance examination for Ritsumeikan university, , 2018.5

The Diophantine equation X3_{=1+9υ over quadratic fields}

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