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I am looking for a set of distinct odd positive integers $3\leq{O_1}<O_2<O_3<...<O_n$ such that

$$\frac{1}{O_{1}}+\frac{1}{O_{2}}+\frac{1}{O_{3}}+...+\frac{m+1}{O_{n}}=1$$ such that $m\in\mathbb{N}$ and $1\leq{m}\leq5$, and other than the simple sets $S=\{3\}$, $m=2$ and $S=\{5\}$, $m=4$.

Is it possible such set of distinct odd positive integers? And could you give me an example?

Thanks in advance!

Juan Moreno
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  • For n=5 $(O_1, O_2, O_3, O_4, O_5)=(3, 3, 5, 9, 45)$ is close to what you mean. I do not think there is other possibility. – sirous Nov 08 '20 at 19:49
  • Thanks @sirous; however, (i) 3 and 3 are not distinct, and (ii) I am not sure if by saying n=5 are you assuming that $n=m$. I also think that an egyptian fraction such as the one I am asking for can not exist, but I am not sure how to prove it, and some example would discard my conjecture – Juan Moreno Nov 08 '20 at 20:22

1 Answers1

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Comment:

You are write and it can be shown by argument. I checked this by following program in Python:

 for a in range (3, 200, 2):

for b in range (3, 200, 2):


 for c in range (5, 200, 2):

   for d in range (9, 200, 2):

     for e in range (11, 200, 2):

        x=a*b*c*d+a*c*d*e+a*b*d*e+b*c*d*e+a*b*c*e

        y=a*b*c*d*e

        z=x/y

        t=int(z)

        if (t*Y)==x:

           print a, b, c, d, e

And only two sets of solutions was resulted"

(a, b, c, d, e)= (3, 3, 5, 9, 45), (3, 3, 5, 15, 15)

If you replace 3 by other odds like 7, 5, 13 . . . in one command started with 3 you will get no result.

sirous
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