No three algebraic conjugates of degree sixteen sum to zero

Let d d be the smallest positive integer, not a multiple of 3 3 , for which there exists an algebraic number α \alpha of degree d d over Q \mathbb {Q} whose three algebraic conjugates add to zero. We prove that d = 20 d=20 . This is derived from the following result: for any linear relation ∑ j = 1 d a j α j = 0 \sum _{j=1}^d a_j \alpha _j=0 with coefficients a j ∈ Z a_j\in \mathbb {Z} among the conjugates α j \alpha _j of an algebraic number of degree d = p m d=p^m , where p p is a prime number, m ⩾ 1 m \geqslant 1 , the sum ∑ j = 1 a j \sum _{j=1}a_j is divisible by p p . If d = 2 p m d=2p^m , p ⩾ 3 p\geqslant 3 and ∑ j = 1 d | a j | > p \sum _{j=1}^d|a_j| > p , then ∑ j = 1 a j \sum _{j=1}a_j is an even number.