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.
Siu Hang Man, Minimal rank of universal lattices and number of indecomposable elements in real multiquadratic fields, Adv. Math. 447 (2024), Paper No. 109694, 38 pp.
Magdaléna Tinková, Bounds on the Pythagoras number and indecomposables in biquadratic fields, Proc. Edinb. Math. Soc. (2) 68 (2025), no. 3, 843-868.
Robin Visser, Database of indecomposable elements over number fields, 2026.
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
