China Mathematical Competition (Complementary Test)

For positive integer $n$, we have
$$\begin{align*} S_{2^{n+1}-1} &= a_1 + (a_2 + a_3) + (a_4 + a_5) + \dots + (a_{2^{n+1}-2} + a_{2^{n+1}-1}) \\ &= u + v + (a_1 + u + a_1 + v) + (a_2 + u + a_2 + v) + \dots + \\ & \qquad (a_{2^{n-1}} + u + a_{2^{n-1}} + v) \\ &= 2^n (u + v) + 2S_{2^{n-1}}. \end{align*}$$
Then
$$\begin{align*} S_{2^{n}-1} &= 2^{n-1}(u+v) + 2S_{2^{n-1}-1} \\ &= 2^{n-1}(u+v) + 2(2^{n-2}(u+v) + 2S_{2^{n-2}-1}) \\ &= 2 \cdot 2^{n-1}(u+v) + 2^2 S_{2^{n-2}-1} \\ &= \dots = (n-1) \cdot 2^{n-1}(u+v) + 2^{n-1}(u+v) \\ &= (u+v) \cdot n \cdot 2^{n-1}. \end{align*}$$
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 $l \equiv k - 1 \pmod 2$. Then $S_{2^{n-1}} = q^2 l^2 \cdot 2^{k-1+q \cdot l^2}$, and
$$\begin{align*} k - 1 + q \cdot l^2 &\equiv k - 1 + l^2 \equiv k - 1 + (k - 1)^2 \\ &= k(k - 1) \equiv 0 \pmod 2. \end{align*}$$
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. $\square$