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!

Your idea of using a sequence with $a_k = 3 a_{k-1}$ seems to work when $a_1$ is even, though the proof doesn't go through.  Try writing an explicit formula for $a_1 + a_2 + \ldots + a_k$ and for $a_1^2 + a_2^2 + \cdots + a_k^2$ - both are geometric series - and then divide one by the other.
When $a_1$ is odd, a geometric series with a different ratio (not 3) will work.
