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A391775
Discriminants of totally real quartic fields in which every additively indecomposable element is a unit.
0
1125, 10512, 15125, 235125, 946125, 1781725, 6670125
OFFSET
1,1
COMMENTS
For any totally real field K, an additively indecomposable element of K is a totally positive element in the maximal order of K which cannot be written as the sum of two totally positive integral elements of K. Here, an element x of K is totally positive if all conjugates of x are positive real numbers. Any totally positive unit is additively indecomposable, although the converse need not hold.
Term a(8) is greater than 10^7. Further terms of this sequence include 88073725 (from Q(sqrt(2523 + 754*sqrt(5)))), 2183005125 (from Q(sqrt(20895 + 8358*sqrt(5)))), and 703556316125 (from Q(sqrt(375115 + 150046*sqrt(5)))).
It is an open problem whether this sequence has infinitely many terms. All known terms correspond to totally real quartic fields of the form Q(sqrt(A + B*sqrt(D))) with D = 3 or 5. A special case of a conjecture of Siu Hang Man [Man 2024, Conjecture 1.7] states that there are only finitely many real biquadratic fields in which every additively indecomposable element is a unit.
LINKS
Martin Čech, Dominik Lachman, Josef Svoboda, Magdaléna Tinková, and Kristýna Zemková, Universal quadratic forms and indecomposables over biquadratic fields, Math. Nachr. 292 (2019), no. 3, 540-555.
Magdaléna Tinková, Bounds on the Pythagoras number and indecomposables in biquadratic fields, Proc. Edinb. Math. Soc. (2) 68 (2025), no. 3, 843-868.
EXAMPLE
a(1) = 1125, as the maximal real subfield K of the cyclotomic field Q(zeta_15) is a totally real quartic field of discriminant 1125 and all additively indecomposable elements of K are units.
The first seven terms arise as discriminants of the following totally real quartic fields:
a(1) = 1125: Q(sqrt(15 + 6*sqrt(5))),
a(2) = 10512: Q(sqrt(11 + 4*sqrt(3))),
a(3) = 15125: Q(sqrt(55 + 22*sqrt(5))),
a(4) = 235125: Q(sqrt(135 + 42*sqrt(5))),
a(5) = 946125: Q(sqrt(435 + 174*sqrt(5))),
a(6) = 1781725: Q(sqrt(363 + 110*sqrt(5))),
a(7) = 6670125: Q(sqrt(1155 + 462*sqrt(5))).
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Robin Visser, Mar 13 2026
STATUS
approved