China Mathematical Competition (Complementary Test)

For positive integer $n$, we have $$\begin{array}{rl}{S}_{2^{n+1}-1}& =a_1+(a_2+a_3)+(a_4+a_5)+\cdots +(a_{2^{n+1}-2}+a_{2^{n+1}-1})\\ & =u+v+(a_1+u+a_1+v)+(a_2+u+a_2+v)+\cdots +\\ & (a_{2^{n-1}}+u+a_{2^{n-1}}+v)\\ & =2^n(u+v)+2S_{2^{n-1}}.\end{array}$$ Then $$\begin{array}{rl}{S}_{2^n-1}& =2^{n-1}(u+v)+2S_{2^{n-1}-1}\\ & =2^{n-1}(u+v)+2(2^{n-2}(u+v)+2S_{2^{n-2}-1})\\ & =2\cdot 2^{n-1}(u+v)+2^2{S}_{2^{n-2}-1}\\ & =\cdots =(n-1)\cdot 2^{n-1}(u+v)+2^{n-1}(u+v)\\ & =(u+v)\cdot n\cdot 2^{n-1}.\end{array}$$ Suppose $u+v=2^k\cdot q$, where $k$ is a non-negative integer, and $q$ is an odd number. Take $n=q\cdot l^2$, where $l$ is any positive integer satisfying $l\equiv k-1(\mathrm{mod}2)$. Then $S_{2^{n-1}}=q^2{l}^2\cdot 2^{k-1+q\cdot l^2}$, and $$\begin{array}{rl}k-1+q\cdot l^2& \equiv k-1+l^2\equiv k-1+(k-1{)}^2\\ & =k(k-1)\equiv 0(\mathrm{mod}2).\end{array}$$ Therefore, $S_{2^{n-1}}$ is a square number. Since there are infinite $l$'s, there are infinite terms in $\{S_n\}$ that are square numbers. The proof is complete. $\square$