All Random Solved Random Open
Let $p:\mathbb{Z}\to \mathbb{Z}$ be a polynomial whose leading coefficient is positive and such that there exists no $d\geq 2$ with $d\mid p(n)$ for all $n\geq 1$. Is it true that, for all sufficiently large $m$, there exist integers $1\leq n_1<\cdots <n_k$ such that \[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}\] and \[m=p(n_1)+\cdots+p(n_k)?\]
Graham [Gr63] has proved this when $p(x)=x$. Cassels [Ca60] has proved that these conditions on the polynomial imply every sufficiently large integer is the sum of $p(n_i)$ with distinct $n_i$. Burr has proved this if $p(x)=x^k$ with $k\geq 1$ and if we allow $n_i=n_j$.