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Statement

Setting: (1) is the equation $a/n=1/x+1/y+1/z$ in positive integers $x,y,z$, repetition allowed; $q,r,s,t$ denote positive integers, and the paper assumes $a>3$ throughout (p. 193). See the Theorem for the setting in full.

Lemma 1 (p. 193). "Suppose that $rn+s\equiv0\ (\mathrm{mod}\ arst-1)$. Then (1) is soluble."

The proof (p. 193) names the solution: if $rn+s+q=arstq$, then $x=stq$, $y=nrtq$, $z=nrst$ solve (1). The congruence supplies such a $q$, namely $q=(rn+s)/(arst-1)$, which is a positive integer.

Source. R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), 193--198, doi:10.1112/S0025579300002886; Lemma 1 and its proof on p. 193. The edition is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image, and the one-line proof was followed: with $rn+s+q=arstq$, $1/(stq)+1/(nrtq)+1/(nrst)=(rn+s+q)/(nrstq)=a/n$. Nothing here is independently reviewed.

Proof pointer

P. 193, one line: the identity above. For $a=4$ the paper points to Chapter 30, § 1 of Mordell's Diophantine equations (1969) for solutions of the same kind.

Dependencies

None.

Bears on

• Problem 242: at $a=4$ the lemma is a sufficient condition for $4/n$ to be a sum of three unit fractions, possibly with repeated denominators: $n$ is representable whenever $rn+s\equiv0\pmod{4rst-1}$ for some positive integers $r,s,t$. It gives a representation for each $n$ in the residue classes it covers and decides no case outside them. The paper uses it, through Lemma 2, to sieve the exceptions in the Theorem.