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
Power integral bases in mixed biquadratic number fields 83 This is only possible if / = 1, 2xo + x 4 = ±1, n\ — 3 (that is ?7i = —3) and x\ = 1. If £4 = 1, then X2 = 0 or x 2 — —1; if £4 = — 1, then x 2 = 0 or x 2 = 1. Then the third factor of the index form is not positive, hence it is equal to —1: Í lA" mi If x 2 = 0, £4 = 1 or x 2 = —1, £4 = 1, then — 3 £3 + — = —1, that is 2 J 4 3(2X*3 -f l) 2 -f- m\ — 4. The first term is 11011-negative, not greater than 4, divisible by 3 and odd, hence it is equal to 3. Then 3(2x'3 -f l) 2 = 3 and m 1 = 1, hence 2X 3 + 1 = ±1» £3 = 0 or X3 = —1. If x 2 = 0, X4 = —1 or x 2 = 1, X4 ~ —1, then —3 (#3 — - ) = —1, that 2 J 4 is 3(2x3 — l) 2 + mi = 4. Similarly as above it follows that the first term is 3 and mi = 1, X3 = 0 or X3 — 1. The remaining cases are (X3, X4) = (0,1), (—1, 1), (0, —1), (1, —1). Considering the second factor we get ^ — ^ = ±1. On the other hand we have / = mi = 1, hence = 0. It means that there are no solutions in case b, either. The cases (c) /(2x2 + X4) 2 = 2, «1X4 = 2, (d) /(2x2 + xa ) 2 — 3, n\x\ = 1, (e) l(2x 2 + ^4) 2 = 4, n xx| = 0 are much simpler to consider. Hence in case 1 there are no power integral bases. Cases 2, 3/A, 3/B are similar to consider. Case J{/A. Now we have mi > 0, r?i < U. Let n 1 = |?2i| > 0. The third factor is non-positive, so multiplying by —2 we get 4/i 1X3 + ?771 (2 x 2 + X4) 2 = 2. The first term is non-negative, less than or equal to 2 and divisible by 4, hence only 4771X3 = 0 and 777.1(2x2 + X4) 2 = 2 are possible. These imply X3 = 0, mi = 2 and 2x 2 + X4 = ±1. Then the second factor is — £4 = —1, that is X4 = ±1. If X4 = 1, then x 2 — 0 or X2 = — 1, and if £4 = — 1, then x 2 = 0 or X2 = 1. The remaining cases are (x'2,X4) = (0,1), (— 1,1), (0, — 1), (1, — 1). Considering the first factor we get i- ^ = ±1. But / > 0, m < 0, so > 0, hence - 1, that Ls l-n x = 2. On the other hand, by l,n 1 £ Z, we get I > 1, ?7i < —1, hence / — n\ > 2. In this inequality the equation holds, so we have / = 1, ?7i = — 1. Summarizing, hi this case / = 1, m\ — 2, ni = — 1 and the solutions of the index form equation are (x 2,x 3 )x 4) = ±(0,0,1), ±(1,0,-1). Case Ji/B. In this case ni > 0, nil < 0, and set mi = |?77i| > 0.
