We start by establishing that for every $k\ge 6$, there is a Pythagorean triple $(k,\ell,m)$ such that $m^2=k^2+\ell^2$, and $\ell>k$. To show this, it suffices to take \[ m-\ell =1 \quad\text{and}\quad m+\ell=k^2 \Leftrightarrow m=\frac{k^2+1}{2}\quad\text{and}\quad \ell = \frac{k^2-1}{2}, \]for $k$ odd; and for $k=2t$ even, \[ m-\ell =2 \quad\text{and}\quad m+\ell =2t^2\Leftrightarrow m=t^2+1\quad\text{and}\quad \ell=t^2-1. \]Having shown this claim, generate now the first triple $(a_1,a_2,a_3)$ where $a_3^2=a_1^2+a_2^2$. Namely, the number $a_3^2$ is a square itself; and moreover, can be expressed as a sum of two squares. Let the next triple we take, whose first coordinate is $a_3$, be $(a_3,a_4,a_5)$, i.e. $a_5^2=a_3^2+a_4^2 = a_1^2+a_2^2+a_4^2$ (note that such a triple exists precisely due to the claim I've shown). Namely, $a_5^2$ can be written as a sum of $1$, $2$, and $3$ distinct squares. Continuing this process and noting that $(a_i)_{i\ge 1}$ is increasing, we are done.