China Mathematical Competition (Complementary Test)

(1) For any $a\in A$, $a=2^k$ ($k\in {\mathbb{N}}^{\ast }$). Then $2a=2^{k+1}$. Let $b$ be any positive integer strictly less than $2a-1$. Then $(b+1)\le 2a-1$.

Between $b$ and $b+1$, one is an odd number that contains no prime factor $2$, and the other is an even number that contains at most the $k$th power of $2$. Therefore, $b(b+1)$ is definitely not a multiple of $2a$.

(2) For $a\in \bar{A}$ and $a\ne 1$, suppose $a=2^{k}m$ where $k$ is a non-negative integer and $m$ is an odd number greater than $1$. Then $2a=2^{k+1}m$. We will present three different proofs in the following.

Proof 1. Let $b=mx$, $b+1=2^{k+1}y$. Eliminating $b$, we have $2^{k+1}y-mx=1$. Since $(2^{k+1},m)=1$, the equation has integral solutions that can be expressed as $$\{\begin{array}{l}x=x_0+2^{k+1}t,\\ y=y_0+mt\end{array}(\text{where }t\in \mathbb{Z},\text{ and }(x_0,y_0)\text{ is a special solution of the equation}.)$$ Denote the smallest solution among them as $(x^{\ast },y^{\ast })$. Then $x^{\ast }<2^{k+1}$. Therefore, $b=m{x}^{\ast }<2a-1$ and $b(b+1)$ is a multiple of $2a$.

Proof 2. Since $(2^{k+1},m)=1$, by the Chinese Remainder Theorem, the congruence equation $$\{\begin{array}{l}x\equiv 0(\mathrm{mod}{2}^{k+1}),\\ x\equiv m-1(\mathrm{mod}m)\end{array}$$ has a solution $x=b$ with $b\in (0,2^{k+1}m)$. It is easy to see that $b<2a-1$ and $b(b+1)$ is a multiple of $2a$.

Proof 3. Since $(2^{k+1},m)=1$, then there exists $r\in {\mathbb{N}}^{\ast }$, $r\le m-1$, such that $2^r\equiv 1(\mathrm{mod}m)$. Take $t\in {\mathbb{N}}^{\ast }$ such that $tr>k+1$. Then $2^{tr}\equiv 1(\mathrm{mod}m)$. It is easy to see that there exists $$b=(2^{tr}-1)-q\cdot 2^{k+1}m>0(q\in \mathbb{N}),$$ such that $0<b<2a-1$. Then we have $m\mid b$, $2^{k+1}\mid b+1$. Therefore, $b(b+1)$ is a multiple of $2a$.