Let $P$ be a regular polygon with $ 4k + 1 $ sides (where $ k $ is a natural) whose vertices are $ A_1, A_2, ..., A_ {4k + 1} $ (in that order ). Each vertex $ A_j $ of $P$ is assigned a natural of the set $ \{1,2, ..., 4k + 1 \} $ such that no two vertices are assigned the same number. On $P$ the following operation is performed: Let $ B_j $ be the midpoint of the side $ A_jA_ {j + 1} $ for $ j = 1,2, ..., 4k + 1 $ (where is consider $ A_ {4k + 2} = A_1 $). If $ a $, $ b $ are the numbers assigned to $ A_ {j} $ and $ A_ {j + 1} $, respectively, the midpoint $ B_j $ is written the number $ 7a-3b $. By doing this with each of the $ 4k + 1 $ sides, the $ 4k + 1 $ vertices initially arranged are erased. We will say that a natural $ m $ is fatal if for all natural $ k $ , no matter how the vertices of $P$ are initially arranged, it is impossible to obtain $ 4k + 1 $ equal numbers through a finite amount of operations from $ m $. a) Determine if the $ 2010 $ is fatal or not. Justify. b) Prove that there are infinite fatal numbers.

HIDE: PS. A help in translation of the 2nd paragraph is welcome. Original wording Diremos que un natural $m$ es fatal si no importa cómo se disponen inicialmente los vértices de ${P}$, es imposible obtener mediante una cantidad finita de operaciones $4k+1$ números iguales a $m$.