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Source. Theorem 1, printed p. 435 (PDF p. 1) of R. L. Graham, A theorem on partitions, J. Austral. Math. Soc. 3 (1963), no. 4, 435--441, DOI 10.1017/S1446788700039045; the Remarks on printed p. 441 (PDF p. 7). The copy read is an image-only scan; the statement and the Remarks were read on the rendered page images.

Statement

Theorem 1 (p. 435). Every integer $n>77$ admits positive integers $k,a_1,\ldots,a_k$ satisfying the three conditions

1. $1<a_1<a_2<\cdots<a_k$;
2. $n=a_1+a_2+\cdots+a_k$;
3. $1=a_1^{-1}+a_2^{-1}+\cdots+a_k^{-1}$.

The Remarks (p. 441) add that the threshold is exact: "in some recent unpublished work of D. H. Lehmer, it has been shown that we must have $r(1,1)\ge77$, i.e., $77$ cannot be partitioned into distinct positive integers whose reciprocals sum to $1$." The same Remarks state the polynomial conjecture $2'$, whose case $\alpha=\beta=1$ is the question of Problem 283 (see the card).

Proof pointer and sketch

The proof (pp. 435--437) is a table of explicit representations for every $n$ from $78$ to $167$ and the odd $n$ from $169$ to $333$ (the entry $n:a_1,\ldots,a_k$ means $\sum a_i=n$ and $\sum1/a_i=1$; the table begins $78:2,6,8,10,12,40$) followed by two transformations of a representation $1=\sum1/d_i$ with denominator sum $U$: $1=\frac12+\sum\frac1{2d_i}$ has denominator sum $2U+2$, and $1=\frac13+\frac17+\frac1{78}+\frac1{91}+\sum\frac1{2d_i}$ has denominator sum $2U+179$; all denominators remain distinct provided no $d_i$ equals $1$ or $39$. The first transformation fills in the even $n$ from $168$ to $334$, and induction then covers every $n>77$. The table was not rechecked here and the proof was read for structure only.

Dependencies and read depth

Self-contained apart from the table. Read depth: claims checked (the statement on the page image of p. 435 and the Remarks on p. 441 were read clause by clause); the proof is not verified here.

Bears on

• Problem 283: the case $p(x)=x$ of the problem's statement, with the explicit threshold $77$ (the site's "Graham [Gr63] has proved this when $p(x)=x$").
• Problem 351: the case $p(x)=x$ without the removal of a finite set. A representation $n=\sum a_i$ with $\sum1/a_i=1$ gives $n+1=\sum(a_i+1/a_i)$, so every integer $m\ge79$ is a finite sum of distinct terms of $\{n+1/n\}$; the strong completeness the problem asks for (any finite set removed) is Theorem 2 of the same paper, or Theorem 3 with $\alpha=1$, and the polynomial case is not treated here.