China Western Mathematical Olympiad

Proof Noting that $S$ is the longest sequence of numbers satisfying the condition. Hence, if we add a term, $0$ or $1$, after the last term of $S$, there will occur two identical sections of successive $5$ terms in $S$, and that is, there exist $i\ne j$ such that $$(a_i,a_{i+1},\dots ,a_{i+4})=(a_{n-3},a_{n-2},\dots ,a_n,0),$$ $$(a_j,a_{j+1},\dots ,a_{j+4})=(a_{n-3},a_{n-2},\dots ,a_n,1).$$ If $(a_1,a_2,a_3,a_4)\ne (a_{n-3},a_{n-2},a_{n-1},a_n)$, then $$1<i,j<n-3,i\ne j,$$ and $$\begin{array}{rl}(a_i,a_{i+1},a_{i+2},a_{i+3})& =(a_j,a_{j+1},a_{j+2},a_{j+3})\\ & =(a_{n-3},a_{n-2},a_{n-1},a_n).\end{array}$$ Now, consider $a_{i-1}$, $a_{j-1}$ and $a_{n-4}$, among which there must be two identical terms. This causes two sections with successive $5$ terms respectively in $S$ to be identical. It leads to a contradiction. Therefore, the proposition holds.