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Statement

A solution $(x,y,z)\in\mathbb N^3$ of $4/n=1/x+1/y+1/z$ is of Type I if $n$ divides $x$ and is coprime to $y,z$, and of Type II if $n$ divides $y,z$ and is coprime to $x$; $f_{\mathrm I}(n)$ and $f_{\mathrm{II}}(n)$ count them (p. 3).

Proposition 1.6 (Vanishing), p. 6, states: "For any odd perfect square $n$, we have $f_{\mathrm I}(n)=f_{\mathrm{II}}(n)=0$."

The paper adds (p. 6) that this does not disprove the Erdős--Straus conjecture, because the inequality $f(n)\ge3f_{\mathrm I}(n)+3f_{\mathrm{II}}(n)$ of (1.2) is not an equality at perfect squares; it attributes the observation essentially to Schinzel and to Yamamoto and notes that a variant appears in Bello-Hernández, Benito and Fernández.

Source. Elsholtz and Tao, arXiv:1107.1010v6, p. 6; read on the page image. Proved in Section 4 (p. 19). Published as J. Aust. Math. Soc. 94 (2013), no. 1, 50--105, DOI 10.1017/S1446788712000468; the published version was not compared.

Read depth. Claims checked: the statement and the paragraph following it were read clause by clause; the proof was not read beyond its outline.

Proof pointer

The proof (p. 19) takes a Type I solution through the parametrization of Proposition 2.2 and reaches a contradiction with quadratic reciprocity and the supplementary laws for the Jacobi symbol; the Type II case is said to be almost identical and is omitted.

Dependencies

The parametrizations of Section 2 (Proposition 2.2 for Type I, with (2.13) and (2.14) for Type II) and quadratic reciprocity, (A.7)--(A.9) of the paper's appendix; not examined here.

Bears on

• Problem 242: for an odd perfect square $n$ every solution, if one exists, is of neither type. The paper draws the consequence (p. 6) that any method showing $f_{\mathrm I}(p)$ or $f_{\mathrm{II}}(p)$ nonzero must fail when the prime $p$ is replaced by an odd square such as $p^2$, which it says rules out strategies such as a finite set of covering congruences. The proposition proves no case of the conjecture and refutes none.