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A combinatorial approach to the \(2/n\) problem. (English) Zbl 0974.11014

This paper gives an elementary approach to the solutions of the Diophantine equation \(\frac{2}{n}= \frac{1}{x}+ \frac{1}{y}\). The authors follow their own path of discovery and treat (step by step) the cases \(n=p\) (prime), \(n=p^r\), \(n=p q\), \(n=pqr\) etc. Let \(\beta_n\) denote the number of solutions of the equation with \(x<y\). Let \(n=p_1^{e_1} \cdots p_k^{e_k}\) denote the prime factorization of an odd integer \(n\). It is shown that \(\beta_n= \frac{1}{2} (2e_1+1)(2e_2+1) \cdots (2e_k+1) - \frac{1}{2}\).
In the concluding remarks it is mentioned that all solutions (for odd \(n\)) can be described by means of the divisors \(u\) of \(n^2\) in the form \(x= \frac{n+u}{2}\) and \(y= \frac{n+v}{2}\), where \(uv=n^2\). It should be pointed out that this classification was used by D. Culpin and D. Griffiths [Egyptian fractions, Math. Gaz. 63, 49-51 (1979; Zbl 0973.11500)] to derive a much shorter proof of the more general formula \(\beta_n= \frac{1}{2} (|2r-1 |\prod_{i=1}^k (2e_i+1) - 1)\), where \(n=2^r p_1^{e_1} \cdots p_k^{e_k}\) For related results on the equation \(\frac{1}{n}= \frac{1}{x}+ \frac{1}{y}\) one should consult (a) E. Sós, Die diophantische Gleichung \(\frac{1}{x}= \frac{1}{x_1}+ \frac{1}{x_2}+ \cdots + \frac{1}{x_n}\), Z. Math.-Natur. Unterricht 36, 97-102 (1905; JFM 36.0271.03), (b) Am. Math. Mon. 103, 171 (Problem 10501) (1996); solution: ibid. 105, 372 (1998).

MSC:

11D68 Rational numbers as sums of fractions
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