Erdős and Graham state that this is open even for $k=2$ and 'the answer should be affirmative but the problem seems very hard'.

Unfortunately the problem is trivially true as written (simply taking $\{1,\ldots,k\}$ and $n>k^{1/\epsilon}$). There are (at least) two possible variants which are non-trivial, and it is not clear which Erdős and Graham meant. Let $P$ be the sequence of $k$ consecutive integers sought for. The potential strengthenings which make this non-trivial are:

• Each $m\in P$ must be $m^\epsilon$-smooth. If this is the problem then the answer is yes, which follows from a result of Balog and Wooley [BaWo98]: for any $\epsilon>0$ and $k\geq 2$ there exist infinitely many $m$ such that $m+1,\ldots,m+k$ are all $m^\epsilon$-smooth.


• Each $m\in P$ must be in $[n/2,n]$ (say). In this case a positive answer also follows from the result of Balog and Wooley [BaWo98] for infinitely many $n$, but the case of all sufficiently large $n$ is open.

See also [370] and [928].

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