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.

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This page was last edited 28 December 2025.