Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2001. Sectio Mathematicae (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 28)

NYUL, G., Power integral bases in mixed biquadratic number fields

82 Gábor Nyúl Theorem. Let K = Q

(\/rn.,^/n) be a mixed biquadratic number field represented in one of the forms listed, above. In cases 1, 2 and 3/A there are no power integral bases. In the other cases the necessary and sufficient condition of the existence of power integral bases in K is Case 3/B: m,\ — —1, I — 4ni = —1 (and by the assumption ni >0/ Case 4/

A: mi = 2, ni = — 1, 1=1, so m — 2 and n = —1. Case 4/B: mi — —2, I — ri\ = ±2 (and by the assumption ni > 0). Case 5/A: ni = — 1, 4

1 — mi = 1 (and by the assumption mi >0^. Case 5/B: I = 1, ni — mi = ±4 (and by the assumption mi, ni < 0J. The solutions of the index form equation corresponding to the above integral basis are Case 3/B (x

2,x

3,x

4) = (1,1, -2), (1,-1, 2), Case 4/A (x

2,x

3, x

4) = (0, 0,1), (1, 0, -1), Case 4/B (x

2,x

3,x4) = (0,0,1), (1,0,-1), Case 5/A m = 3, n =-1 (i

2, ®

3, x

4) = (1, -2,1), (1, -2,-1), (0,1,0), (1, —1,0), Case 5/A other fields (x

2,x

3,x±)= (1, — 2,1), (1, — 2, — 1), Ca.se 5/B (x

2,x

3,x

4) = (0,1, 0), (1,-1, 0). Note that if (x

2, x

3, is a solution then so also is (—x

2, — x

3

) — £4) but we include only one of them. Proof of the Theorem. In each case we solve equation (1) using the relevant index form. In each case the index form splits into three factors taking integer values, hence all factors must be equal to ±1. We detail some tipical cases, the others are similar to deal with. Case 1. We have mi > 0, n\ < 0, mi=l (mod 4), ni=l (mod 4). Set n

1 = |mI > 0. The first factor of the index form is non-negative. Multiplying by 4 we get /(2x2 + ÍC4)

2 + = 4. On the left hand side both terms are non-negative integers, hence we have to consider the following five cases (a. to e.): (a) /(2x2 + x±)

2 = 0, n\x\ = 4 By I > 0, we have 2x

2 + X4 = 0, that is 2 | X4. On the other hand («1,^4) = (1,4), (4,1), and X4 is even which imply rii = 1, that is n 1 = —1. Then we obtain ni^l (mod 4), a contradiction, hence in case a there are 110 solutions. (b) /(2x

2 + Z4)

2 = 1, n

xx\ = 3