Question 85
Let $A = \{2, 2^2, \dots, 2^n, \dots\}$. Prove:
(1) For any $a \in A$, $b \in \mathbb{N}^*$, if $b \lt 2a - 1$, then $b(b+1)$ will not be a multiple of $2a$.
(2) For any $a \in \bar{A} (= \mathbb{N}^* - A)$ satisfying $a \neq 1$, there exists $b \in \mathbb{N}^*$ satisfying $b \lt 2a - 1$, such that $b(b+1)$ is a multiple of $2a$.
(1) For any $a \in A$, $b \in \mathbb{N}^*$, if $b \lt 2a - 1$, then $b(b+1)$ will not be a multiple of $2a$.
(2) For any $a \in \bar{A} (= \mathbb{N}^* - A)$ satisfying $a \neq 1$, there exists $b \in \mathbb{N}^*$ satisfying $b \lt 2a - 1$, such that $b(b+1)$ is a multiple of $2a$.