Determine all sequences $a_1, a_2, a_3, . . . $ of nonnegative integers such that $a_1 \lt a_2 \lt a_3 \lt · · ·$ Determine all sequences $a_1, a_2, a_3, . . . $ of nonnegative integers such that $a_1 \lt a_2 \lt a_3 \lt · · ·$
and $a_n$ divides $a_{n-1}+n$ for all $n\ge2$.
I know that one obvious possible sequence is $a_n=a_{n-1}+n$ but I don't know how to prove this is the only one or if there is more
from the 2018 SAMO senior round 3
http://www.samf.ac.za/content/files/QuestionPapers/s3q2018.pdf

I will try and prove that $a_n = a_{n-1} + n$ isn't the only solution.
$a_n$ divides $a_{n-1}+n$, So we can take 'm' to be the quotient. Note that m is an integer.
This gives us $ma_n=a_{n-1}+n$
Putting n=2 we get,  $ma_2 = a_1 + 2$ 
As  $a_2>a_1$ and both of them being being integers, $a_2-a_1\ge1$
$a_2\ge a_1+1$
$a_2 +1\ge a_1+2$
$a_2 +1\ge ma_2$
$(m-1)a_2\le 1$
$m\le {a_2+1\over a_2}$, Note that $a_2\ge 1$.
Trying out any value of $a_2$, we get,  $m\le 2$ and because $m$ is an integer, $m=1$ or $m=2$.
Which give us $a_n=a_{n-1}+n$ or $2{a_n}=a_{n-1}+n$
Solve the first equation by telescopy and I don't know how to solve the second equation.
For first equation you will get $a_n= a_1 -1 + {n^{2}+n \over 2}$.You can take $a_1$ to be any non-negative integer. Try solving equation 2. 
