Problem Statement
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\[m=p(n_1)+\cdots+p(n_k)?\]
Categories:
Number Theory Unit Fractions
Progress
Graham [Gr63] has proved this when $p(x)=x$. Graham also conjectures that this remains true with $1$ replaced by an arbitrary rational $\alpha>0$ (provided $m$ is taken sufficiently large depending on $\alpha$).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 $n_i=n_j$.
Alekseyev [Al19] has proved this when $p(x)=x^2$, for all $m>8542$. For example,\[1=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\]and\[200 = 2^2+4^2+6^2+12^2.\]van Doorn [vD25] has investigated the question of what 'sufficiently large' means for $p(x)=x$. van Doorn has also proved the original conjecture for many linear and quadratic polynomials, for example $p(x)=x+5$ or $p(x)=x^2+100$ - see the comments section.
Source: erdosproblems.com/283 | Last verified: January 14, 2026