Say $b_{n}=2a_{n}-a_{n-1}$. Then $b_{i}\not\equiv b_{j}\pmod{n}$ for all $n \geq i,j$, hence $|b_{i}-b_{j}|<\max(i,j)$. Note that $b_{1}=2$, so $b_{n}\leq n+1$ holds, which says $a_{n}\leq n$. Suppose that there is a $k$ that $b_{k}<k+1$, then $a_{n}<n$ for all $n>k$, which contradicts (3). Hence $b_{n}=n+1$ for all $n$, so $a_{n}=n$ is the unique solution.

Generalization: Suppose $\{a_n\}_{n\ge 0}$ satisfies the following conditions. (1) $a_n\in \mathbb{Z}$ (2) $a_0=0, a_1=1$ (3) For every $n\ge2$, $\{2a_i-a_{i-1} | i=1, 2, 3, \cdots , n\}\equiv \{0, 1, 2, \cdots , n-1\}$ $\mod n$ Is it true that $\{a_n\}_{n\ge 0}$ is one of these three solutions? - $a_{n}=n(n\geq 0)$ - $a_0=0,a_1=1,a_n=n-1(n\geq 2)$ - $a_0=0,a_1=1,a_2=1,a_n=5-n(n\geq 3)$