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

84 Gábor Nyúl Now the second factor is not negative, hence it is equal to 1. If we multiply it by 2, we get 41 x $ + rhix± = 2. On the left hand side the first term is equal to 0, because it is non-negative, not greater than 2 and divisible by 4, which implies ilx^ = 0, mi £4 — 2. From these we get X3 = 0, m\ — 2 (that is mi = —2), x± = ±1. Then the third factor is (2x

2 + x*4)

2 = 1, hence 2xo + £4 = ±1. If £4 = 1, then X2 = 0 or X2 = —1; if £4 = —1, then £

2 = 0 or X2 — 1. In the remaining cases ((x

2,z

4) = (0,1), (-1,1), (0,-1), (1,-1)) the first factor is \ - ^ = ±1, that is I — n\ = ±2. Summarizing, we get m\ — —2, I — ri\ — ±2 and the solutions of the index form equation in this case are (x*

2, £4) = ±(0, 0,1), ±(1, 0, —1). Cases 5/A, 5/B can be discussed in a similar way. In each case it is simple to verify by substitution that the triples (a,*

2, X3, X4) obtained above are indeed solutions of the index form equation. 3. Description of the table We present a list of till mixed biquadratic fields up to discriminant 10

4. In this table D/i, mj{

, // denote the discriminant, the field index and the minimal index, respectively. They are followed by the solutions of I(x

2,23,24) = ±/i, that is the coordinates of the elements of minimal index. If (x'

2,x'3,a;4) is a solution then so also is (—X2, —X3, —X4) but we list only one of them. To construct the table we used [6] (integral basis, Dk), [2] (to calculate mIn order to determine the minimal index ß we took the multiplies K • 'IN.A of TTIK until the index form equation with right hand side ±k • rriK had solutions. In [3] the authors provided a similar list of totally real biquadratic fields. These computations were performed in MAPLE and took just a few minutes.