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How to prove or disprove that there exists a set of natural numbers $$a_{1}<a_{2}<\cdots$$ with the property that, for every $i<j$, $$a_{j}-a_{i}\mid a_{j}-1\quad ?$$ I think it doesn't work for very big $a_{j}$'s and very small $a_{i}$'s.

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up vote 2 down vote accepted

There is no such sequence.

Suppose by contradiction that we have such a sequence, and for any $j>i$ we have $a_j-a_i\vert a_j-1$. Then either $a_j=1$ or $a_i=1$ or $a_j>2(a_j-a_i)$ (because $a_j=1+k(a_j-a_i)$ for some $k\geq 0$, the three cases corresponding to $k=0,1$ and $k\geq 2$).

The first two cases are only possible for small $i$, so we can discard them (we can assume that the sequence starts with arbitrarily large numbers by cutting off the beginning).

The last one gives us $2a_i>a_j$ with any sufficiently large fixed $i$ and all $j>i$. But of course $a_{i+a_i}\geq 2a_i$, so we have a contradiction.

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