Show that there exists an infinite sequence of positive integers $a_1, a_2, \cdots$ such that $$S_n=a_1^2 + a_2^2 + \cdots + a_n^2$$ is square for any positive integer $n$.
Let's first observe the following recursive solutions: $$3^2 + 4^2 = 5^2$$ $$3^2 + 4^2+12^2 = 13^2$$ $$3^2 + 4^2 +12^2 + 84^2= 85^2$$
in which the left side of the previous relation is used to construct the next relation. Therefore, the given problem is equivalent to show that with a known integer $p$, the equation $p^2 + m^2 = n^2$ is always solvable. For example, the second relation above is equivalent to showing that $5^2 + m^2 = n^2$ is solveable, and the third relation is equivalent to showing $13^2 + m^2 = n^2$ is solvable.
$$p^2 + m^2 = n^2 \Leftrightarrow p^2 = n^2 - m^2 = (n+m)(n-m)$$
To solve this equation, we abtrarily let $n-m=1$. Then we have $$n=\frac{p^2+1}{2}\quad\text{and}\quad\frac{p^2-1}{2}$$
$$p^2 + m^2 = n^2 \Leftrightarrow p^2 = n^2 - m^2 = (n+m)(n-m)$$
To solve this equation, we abtrarily let $n-m=1$. Then we have $$n=\frac{p^2+1}{2}\quad\text{and}\quad\frac{p^2-1}{2}$$
Now, it is clear that as long as $p$ is odd, both $m$ and $n$ will be integer. Further more, if $p$ is odd, $n$ will be odd too (why?) This means, as long as we have an initial odd $p$, we can get an odd $n$. And because $n$ will become $p$ in the next iteration, this process can continue forever. In another word, the original claim will hold.
Clearly, we have an initial odd $p = 5$. Therefore the claim holds.