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Statement

Setting: (1) is the equation $a/n=1/x+1/y+1/z$ in positive integers, $p$ denotes a prime and $a>3$ (p. 193). The paper's (3) and (4) (p. 193) define

$$f_1(p)=\begin{cases}\dfrac12\displaystyle\sum_{t\mid(p+1)/a}|\mu(t)|\,d\Bigl(\frac{p+1}{at}\Bigr)&(p\equiv-1\ (\mathrm{mod}\ a)),\\[2ex]0&(\text{otherwise}),\end{cases} \qquad f(p)=[f_1(p)],$$

with $\mu$ the Möbius function, $d$ the number of divisors and $[\,\cdot\,]$ the integer part.

Lemma 2 (p. 194). "For each prime $p$ there are at least $f(p)$ residue classes modulo $p$ so that, if $n$ is a member of any one of them, (1) is soluble."

Source. R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), 193--198, doi:10.1112/S0025579300002886; the definitions (3) and (4) on p. 193, Lemma 2 and its proof on p. 194. The edition is identified on the source card.

Read depth. Claims checked: the statement and the definitions (3)--(4) were read clause by clause on the page images, and the proof was followed. Nothing here is independently reviewed.

Proof pointer

P. 194. Only $a\mid p+1$ needs proof, since otherwise $f(p)=0$. Each triple $(r,s,t)$ of positive integers with $arst=p+1$, $t$ squarefree and $s\le((p+1)/(at))^{1/2}$ gives the class $rn+s\equiv0\pmod p$, on which (1) is soluble by Lemma 1 because $arst-1=p$. Two such triples giving the same class satisfy $s_1^2t_1\equiv s_2^2t_2\pmod p$ with both sides below $p$, hence are equal, and squarefreeness of the $t_i$ then forces the triples to coincide. The paper then says the lemma follows from Lemma 1, (4) and (3); the count, spelled out here, is that for each squarefree $t$ the factorisations $rs=(p+1)/(at)$ with $s\le\sqrt{rs}$ number at least half of $d((p+1)/(at))$, so the triples number at least $f_1(p)\ge f(p)$.

Dependencies

Lemma 1 and the definitions (3)--(4).

Bears on

• Problem 242: at $a=4$ and for a prime $p\equiv3\pmod4$, the lemma names at least $f(p)$ residue classes modulo $p$ on whose members $4/n$ is a sum of three unit fractions, possibly with repeated denominators. These are the classes removed by the large sieve in the proof of the Theorem; the lemma decides no single $n$ outside them.