# $n^2$ divides $n$-th partial sum

Prove or disprove existence of sequence that satisfies:

1. $a_i\neq a_j, i\neq j$,
2. $\{a_i\}_{i=1}^{\infty}=\mathbb{N}$,
3. $n^2\mid\sum_{i=1}^n a_i$ for all $n$

Odd numbers could satisfy the third condition, but it isn't bijecton onto $\mathbb{N}$, obviously.

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Suppose you have started a sequence $a_1, \ldots, a_n$ where $i \neq j \implies a_i \neq a_j$ and $k^2 | \sum_{i=1}^k a_i$.

Pick any $a \in \mathbb{N} - \{a_1 \ldots a_n\}$. Say you want to find some a suitable integer $a_{n+1}$ such that the sequence continued by $a_{n+2} = a$ works. Note $s = \sum_{i=0}^n a_n$.

You need an $a_{n+1}$ satisfying $s+a_{n+1} \equiv 0 \pmod{(n+1)^2}$ and $s + a_{n+1} + a \equiv 0 \pmod{(n+2)^2}$. Write $s+a_{n+1} = k(n+1)^2$. You get the condition $k(n+1)^2+a \equiv 0 \pmod {(n+2)^2}$.

Since $(n+1)$ and $(n+2)$ are coprime, $(n+1)$ is invertible in $\mathbb{Z}/(n+2)^2\mathbb{Z}$, and this condition is equivalent to $k \equiv -a/(n+1)^2 \pmod{(n+2)^2}$.

So there exists infinitely many suitable $k$, and thus infinitely many possibles values for $k(n+1)^2-s = a_{n+1}$. By choosing $k$ large enough, you can choose $a_{n+1}$ larger than $a$ and than any $a_i$ so that the sequence is still injective.

Therefore, you can extend any given sequence $a_1 \ldots a_n$ into a larger sequence $a_1 \ldots a_{n+2}$ where $a_{n+2}$ is any number you want. Repeat the process to fill all the holes (for example by systematically choosing $a_{n+2} = \min (\mathbb{N} - { a_1 \ldots a_n }$) and obtain a complete sequence satisfying your conditions.

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