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Let $n$ be a fixed natural number. How to solve the following equation in natural numbers: $$ \frac{1}{x_1} + \frac{2}{x_2} + \cdots + \frac{n}{x_n} = 1 $$ (I can find many soltions but I am looking for all solutions)

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  • $\begingroup$ What are $x_1...$ integers or natural numbers $\endgroup$ – Archis Welankar Jan 5 '16 at 5:21
  • $\begingroup$ @ArchisWelankar natural numbers $\endgroup$ – alex alexeq Jan 5 '16 at 5:21
  • $\begingroup$ It is sufficient to solve the equation. $$\frac{1}{x_1}+\frac{2}{x_2}+\frac{3}{x_3}=\frac{a}{b}$$ Number $a,b - $ will be set. Which are obtained if we ask the other numbers at their discretion. $\endgroup$ – individ Jan 5 '16 at 5:32
  • $\begingroup$ You can use this formula. math.stackexchange.com/questions/450280/… $\endgroup$ – individ Jan 5 '16 at 5:44
  • $\begingroup$ I see what you mean that it's trivial to write down some solutions e.g. $x_i=in$. But do you have a good reason to believe it's easy to characterize all solutions? That seems like it may be intractable. $\endgroup$ – Gregory Grant Jan 5 '16 at 6:06
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We assume x1 < x2/2 < x3/3 ... and We may write 1/(x1/1) +1/(x2/2) +1/(x3/3) +. . . 1/(xn/n)=1 Number of terms is n then 1/(x1/1)<1/n. taking x1 an arbitrary value satisfying 1/(x1) < 1/n and finding equation:

1/(x2/2) +1/(x3/3) +. . . 1(xn/n)=1-1/x1

Now 1/(x2/2) < 1-1/x1 defines lowest value and also n-1/(x2/2) > 1-1/x1 defines upper value of x2. we take a value in this range. In this way we can find xi; i=3, 4, 5 ... .

Example: 1/2 + 2/6 + 3/21 + 4/172 + 5/9030 = 1

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