Let $\epsilon>0$ and $k\geq 2$. Is it true that, for all sufficiently large $n$, there is a sequence of $k$ consecutive integers in $\{1,\ldots,n\}$ all of which are $n^\epsilon$-smooth?

Let $\epsilon>0$ and $k\geq 2$. Is it true that, for all sufficiently large $n$, there is a sequence of $k$ consecutive integers in $\{1,\ldots,n\}$ all of which are $n^\epsilon$-smooth?

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. We formalize the second: each $m\in P$ (where $P$ is the sequence of $k$ consecutive integers sought for) must be in $[n/2,n]$. In this case a positive answer also follows directly from the result of Balog and Wooley [BaWo98] for infinitely many $n$. Proving this is true for all large $n$ does not follow immediately from [BaWo98], but can be deduced using a similar construction, as shown by SkyYang.