View of The Diophantine system $x^2-6y^2=-5,\ x=2z^2-1$

{orders}ms/98711/cohn.3d -17.11.00 - 12:37

THE DIOPHANTINE SYSTEM x

ÿ 6y

ˆ ÿ5; x ˆ 2z

ÿ 1

J. H. E. COHN^*

1. Introduction.

In a recent paper [5] showing that the system of the title has only the solu- tions in integers given by zˆ0;1;2;3;6 and 91, the introduction men- tioned in passing that the elementary algebraic methods developed for ex- ample in [2] do not appear su¤cient to solve this problem. The authors then proceeded to prove their result using the high powered analytical methods for which they, Tzanakis, de Weger, Steiner and others have become re- nowned. We present below a short simple proof of their result.

It is clearly su¤cient to assume thatx ÿ1;y1 and z0 we shall do so without further mention. The equation v²ÿ6u²ˆ1 has fundamental so- lution ˆ5‡2 

p6

, and then it is easily shown that the general solution of x²ÿ6y²ˆ ÿ5 is given by the two classes x‡y 

p6

ˆ …1‡ 

p6

†ⁿ with n0. Let ˆ5ÿ2 

p6

and de¢ne the sequences unˆ …ⁿÿⁿ†=…ÿ†, v_n ˆ …ⁿ‡ⁿ†=2. Then xˆ v_n‡12u_n, and so our problem reduces to proving that the only solutions of the equation

2z²ˆ1‡v_n‡12u_n …1†

are given bynˆ0;1 and 4 of the equation 2z²ˆ1ÿvn‡12un

…2†

are given bynˆ0;1 and 2.

Here bothunandvnsatisfy the recurrence relationwn‡2ˆ10wn‡1ÿwnand the ¢rst few values are given in the following table:

MATH. SCAND. 82 (1998), 161^164

Received January 9, 1996.

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n un vn vn‡12un ÿvn‡12un

0 0 1 1 í-1

1 1 5 17 7

2 10 49 169 71

3 99 485 1673 703

4 980 4801 16561 6959

We shall quote several standard identities involving the sequencesun and v_n and results concerning the periodicity of these sequences modulo certain moduli which are easily veri¢ed, without writing them all out in detail.

2. Preliminaries.

Lemma 1. The equation 3x⁴ÿ2y²ˆ1 has only the solutions in non-negative integers given by xˆ1;3.

This is shown in [1].

Lemma2. The equation 6y²ˆx⁴ÿ1has only the solutions in non-negative integers given by xˆ1;7.

This is shown in [3].

Lemma3. The equation y² ˆ24x⁴‡1has only the solutions in non-negative integers given by xˆ0;1.

For,…y‡1†…yÿ1† ˆ24x⁴ and here…y‡1;yÿ1† ˆ2. Thus withxˆx1x2

either y1ˆ2x⁴₁;y1ˆ12x⁴₂ in which case1ˆx⁴₁ÿ6x⁴₂; here the lower sign is impossible modulo 3, whereas the upper one gives onlyx₁ˆ1;x₂ˆ0 by Lemma 2;

or y1ˆ6x⁴₁;y1ˆ4x⁴₂and now1ˆ3x⁴₁ÿ2x⁴₂. Again the lower sign is impossible modulo 3, and the upper sign gives only x1ˆx2ˆ1 in view of Lemma 2.

Lemma4. The equation3y²ˆ2x⁴‡1has only the solution in non-negative integers given by xˆ1.

This is proved in [4], and is the only result that we use which has not been proved by technically elementary methods.

162 j. h. e. cohn

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3. Proof of result.

There are four cases.

(A) The only solutions of (1) withn even arenˆ0 and 4.

For with nˆ2k;2z²ˆ1‡v_2k‡12u_2kˆ2v_k…v_k‡12u_k†, and so since …v_k;v_k‡12u_k† ˆ1 we must havev_kˆz²₁. Then 1ˆv²_kÿ24u²_kˆz⁴₁ÿ6…2u_k†², and so by Lemma 2,z₁ ˆ1 or 7 whencekˆ0 or 2 as required.

(B) The only solution of (1) withnodd isnˆ1.

For withnˆ2k‡1,

2z²ˆ1‡v_2k‡1‡12u_2k‡1

ˆ1‡ …5v2k‡24u2k† ‡12…v2k‡5u2k†

ˆ …v²_kÿ24u²_k† ‡17…v²_k‡24u²_k† ‡168ukvk

ˆ18v²_k‡168ukvk‡384u²_k

ˆ6…v_k‡4u_k†…3v_k‡16u_k†:

Here the ¢nal two factors on the right have no common factor, and since 3vk‡16uk3 (mod 4†, we must havevk‡4ukˆz²₁;3vk‡16ukˆ3z²₂. Thus 4ukˆ3z²₂ÿ3z²₁;vkˆ4z²₁ÿ3z²₂, and so

1ˆv²_kÿ24u²_kˆ …4z²₁ÿ3z²₂†²ÿ²⁷₂ …z²₁ÿz²₂†²; whence 2ˆ5z⁴₁‡6z²₁z²₂ÿ9z⁴₂ˆ6z⁴₁ÿ …z²₁ÿ3z²₂†²; or 1ˆ3z⁴₁ÿ2…¹₂…z²₁ÿ3z²₂††²;

and so by Lemma 1, the only possibilities arez₁ ˆ1 or 3. The former gives kˆ0 whencenˆ1 and the latter no solution.

(C) The only solutions of (2) withn even arenˆ0 and 2.

For with nˆ2k;2z²ˆ1ÿv_2k‡12u_2kˆ24u_k…ÿ2u_k‡v_k†, and so since v_kÿ2u_kis odd and has no factor in common withu_kwe must have

either ukˆz²₁;vkÿ2ukˆ3z²₂; in this case 1ˆv²_kÿ24u²_kˆv²_kÿ24z⁴₁ implies z1 ˆ0 or 1 in view of Lemma 3; the latter yields nˆ2 whilst the former gives no solution;

or u_kˆ3z²₁;v_kÿ2u_kˆz²₂. Sinceu_k0 (mod 3), it follows that 3jk, and with kˆ3m we ¢nd that z²₁ˆ¹₃u_3mˆu_m…32u²_m‡1†. This implies that u_m is a square, and just as above this is possible onlyumif equals 0 or 1. The former givesnˆ0 and the latter no solution.

(D) The only solution of (2) withnodd isnˆ1.

For withnˆ2k‡1,

the diophantine system ... 163

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2z²ˆ1ÿv2k‡1‡12u2k‡1

ˆ1ÿ …5v2k‡24u2k† ‡12…v2k‡5u2k†

ˆ …v²_kÿ24u²_k† ‡7…v²_k‡24u²_k† ‡72ukvk

ˆ8…v_k‡3u_k†…v_k‡6u_k† and this is possible only ifz²₁ˆvk‡6uk. But then

2z⁴₁‡1ˆ2…vk‡6uk†²‡1

ˆ2v²_k‡24ukvk‡72u²_k‡v²_kÿ24u²_k

ˆ3…vk‡4uk†²

and by Lemma 4, this is possible only forkˆ0.

This concludes the proof.

Note added in proof. By an extraordinary coincidence this same pro- blem was considered by R. J. Stroeker & B. M. M. de Weger in ``On a quartic Diophantine Equation'', Proc. Edinburgh Math. Soc. 39 (1996), 97^

114. Their method is as long and similar in conception to that of [5] although the details are totally di¡erent.

REFERENCES

1. Richard T. Bumby,The Diophantine equation3x⁴ÿ2y²ˆ1, Math. Scand. 21 (1967), 144^

2. J. H. E. Cohn,148. Some quartic Diophantine equations, Paci¢c J. Math. 26 (1968), 233^243.

3. J. H. E. Cohn,The Diophantine equation x⁴ÿDy²ˆ1, Quart. J. Math. Oxford (2) 26 (1975), 279^281.

4. Wilhelm Ljunggren,U«ber die unbestimmte Gleichung Ax²ÿBy⁴ˆC, Arch. for Mathematik og Naturvidenskab (41) Nr. 10.

5. Maurice Mignotte & Attila Pethoë, On the system of Diophantine Equations x²ÿ6y²ˆ ÿ5;xˆ2z²ÿ1, Math. Scand 76 (1995), 50^60.

DEPARTMENT OF MATHEMATICS

ROYAL HOLLOWAY UNIVERSITY OF LONDON EGHAM, SURREY TW20 0EX

U.K.

e-Mail J.Cohn@rhbnc.ac.uk

164 j. h. e. cohn