It is known that the Erdos-Straus conjecture is about writing $4/n$ as three unit fractions. My question is whether it is known that if $a>4$ $$ \frac an=\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_k} $$ where $k<a$? Or it is a conjecture similar to Erdos-Straus one with the same hardness? For instance, is it known whether $$ \frac 5n=\frac1{x_1}+\frac1{x_2}+\frac1{x_3}+\frac1{x_4} ? $$
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1$\begingroup$ This is an open problem, as Elsholtz comments in ams.org/journals/tran/2001-353-08/S0002-9947-01-02782-9/… (see the second para of the second page of the article). $\endgroup$– LuciaCommented Aug 3, 2015 at 3:32
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$\begingroup$ Can be reduced to 3 term. And then use this formula. math.stackexchange.com/questions/450280/… $\endgroup$– individCommented Aug 3, 2015 at 4:51
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$\begingroup$ See also D11 in Guy, Unsolved Problems In Number Theory, 3rd edition. $\endgroup$– Gerry MyersonCommented Aug 3, 2015 at 5:39
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$\begingroup$ Decomposition into fractions $$\frac{5}{n}=\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$ Always possible - because can take. $x_4=n$ Which leads to the well-known formula. $\endgroup$– individCommented Aug 4, 2015 at 4:44
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$\begingroup$ For the case of $\frac{5}{n}$ see http://math.stackexchange.com/questions/56909/sierpinskis-conjecture. $\endgroup$– abxCommented Aug 4, 2015 at 9:34
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2 Answers
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There is a conjecture for 5/n identical to the Erdos-Straus conjecture and even for k/n, k any positive integer, see the Wikipedia article on the Erdos-Straus conjecture for details. Frank Okoh
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This is a very partial answer. If $n$ is a practical number, then for every $a<n$ there is a reduction of $a/n$ as $$\frac an=\frac1{x_1}+\dots\frac1{x_k}$$ with $k<a$.
