Given $a_1$, find an increasing sequence so that $a_1+\dots+a_k $ divides $ a_1^2+\dots+a_k^2$ for all $k$ Prove that for every natural number $a_1>1$ there's an infinite series $a_1<a_2<a_3<...$ such that for every natural number $k,$ $a_1+a_2+...+a_k \vert a_1^2+a_2^2+...+a_k^2$.
At first glance, I thought this problem is an induction problem and could be solved by forming the sequence inductively. I tried that and didn't get anywhere even though I'm pretty sure there must be an inductive solution.
I went in a different direction, experimenting with series and trying the series
$$a_2=3\cdot a_1$$
$$a_3=3\cdot a_2=9\cdot a_1$$
$$a_4=3\cdot a_3=9\cdot a_2=27\cdot a_1$$
and so on... (esentially every number is three times the number before it in the series)
This way, if we assume the condition stands for $k-2$ and try to prove it for $k-1$ we get
$$a_1\cdot (\frac{3^k-1}{2}) \vert a_1^2\cdot (\frac{9^k-1}{8})$$
From here, we know $a_1 \vert a_1^2$ so we need $\frac{3^k-1}{2} \vert \frac{9^k-1}{8}$, but since $\frac{9^k-1}{8}=\frac{3^{2k}-1}{8}=\frac{(3^k-1)(3^k+1)}{8}=\frac{(3^k-1)}{2}\cdot \frac{(3^k+1)}{4}$, so actually we get that $\frac{3^k-1}{2} \vert \frac{9^k-1}{8}$ if $\frac{(3^k+1)}{4}$ is a whole number. This only works for $k\equiv 1$ (mod 2). I'm not sure where to go from here, I've tried proving this, but for even numbers and haven't gotten anywhere. I'm worried I might be going deep down the series solution rabbithole even though it might not lead anywhere. Any help is appreciated, thanks!

We're trying to find a strictly increasing set of integers $a_i$, with $a_1 \ge 2$, so for all $k \ge 1$
$$\sum_{i=1}^{k}a_i \, \mid \, \sum_{i=1}^{k}a_i^2 \tag{1}\label{eq1A}$$
Unfortunately, I don't know of any way to finish what you've tried. Instead, as you surmised, there's an inductive solution. For $k = 1$, \eqref{eq1A} is true since $a_1 \mid a_1^2$. Assume that, for some $m \ge 1$, \eqref{eq1A} is true is for $k = m$. Set
$$j = \sum_{i=1}^{m}a_i \tag{2}\label{eq2A}$$
Thus, by \eqref{eq1A}, since $a_1 \ge 2$, we have
$$\sum_{i=1}^{m}a_i^2 = jn, \; \; n \ge 2 \tag{3}\label{eq3A}$$
Let
$$a_{m+1} = j(n + j - 1) \tag{4}\label{eq4A}$$
Since $n \ge 2$ and $j \ge 2$, then \eqref{eq4A} and \eqref{eq2A} give that $a_{m+1} \gt j \; \; \to \; \; a_{m+1} \gt a_{m}$. Using $k = m + 1$ in \eqref{eq1A}, the LHS becomes
$$\sum_{i=1}^{m+1}a_i = j + j(n + j - 1) = j(n + j) \tag{5}\label{eq5A}$$
The RHS of \eqref{eq1A} is then
$$\begin{equation}\begin{aligned}
\sum_{i=1}^{m+1}a_i^2 & = jn + [j(n + j - 1)]^2 \\
& = j(n + j\,[(n+j) - 1]^2) \\
& = j(n + j\,[(n+j)^2 - 2(n+j) + 1]) \\
& = j(n + j\,[n+j][n+j-2] + j) \\
& = j(n+j)[1 + j(n+j-2)]
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
From \eqref{eq5A}, the LHS divides the RHS of \eqref{eq1A}, so it's true also for $k = m + 1$. Thus, by induction, we have \eqref{eq1A} is true for all $k \ge 1$.
