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Consider the Diophantine equation $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1$$ I'll give you the solutions sets $(a_1,a_2,\ldots,a_n)$ for $n=1$ up to $4$, because beyond $n=4$ the number of solutions seem to explode. (These can be found quite easily by considering the smallest value of the $n$ variables and seeing just how large/small it can be, and then doing the same for the next smallest variable, and so on.) I will not consider permutations of each set as a separate solution set. $$\begin{align} n=1:&~~~(1)\\ n=2:&~~~(2,2)\\ n=3:&~~~(3,3,3)~(2,3,6)~(2,4,4)\\ n=4:&~~~(4,4,4,4)~(3,3,4,12)~(3,3,6,6)~(3,4,4,6)~(2,3,7,42)\\&~~~(2,3,8,24) ~(2,3,9,18)~(2,3,10,15)~(2,3,12,12)~(2,4,5,20)\\ &~~~(2,4,6,12)~(2,4,8,8)~(2,5,5,10)~(2,6,6,6) \end{align} $$ Denote by $a_n$ the number of solution sets we have for the Diophantine equation above for $n$ variables. So we have $$a_n=1,~~a_2=1,~~a_3=3,~~a_4=14$$ (Thanks for the correction @player3236). And I suppose you could include $a_0=0$ if you really wanted, although I'm not sure how much sense that makes. My question is:

Is there a formula that tells us the number of solutions sets for the Diophantine equation above? If so, how can I find and prove it?

Thank you for your help.

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    $\begingroup$ You missed $(3,4,4,6)$. See oeis.org/A002966. It has keywords hard, more, which says that we don't know the exact values for $n \ge 9$. $\endgroup$
    – player3236
    Commented Feb 14, 2021 at 12:44
  • $\begingroup$ @player3236 thanks for the correction and link. So it's an unsolved problem in number theory? $\endgroup$
    – A-Level Student
    Commented Feb 14, 2021 at 12:50
  • $\begingroup$ An observation I made: Consider $f(x) = 1/x$, this function has $f''>0$ for $x>0$, then it is clear that by Jensen's inequality: $$f\left(\frac{x_1+x_2+\cdots+x_n}n\right)\leq\frac{f(x_1)+f(x_2)+\cdots+f(x_n)}n$$ or $$\frac{n^2}{x_1 + x_2..+x_n}\leq\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}=1$$ Then it is clear that: $$n^2 \leq x_1+x_2+\cdots+x_n$$ Suppose we order the the integers, $x_1\leq x_2\leq\cdots\leq x_n$, the above sum must be less than $n\times x_n$: $$n\leq x_n$$ That gives a lower bound on $x_n$. $\endgroup$
    – Cathartic Encephalopathy
    Commented Feb 14, 2021 at 13:07
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    $\begingroup$ @Buraian Isn't that also immediate from the contrapositive? If $x_n<n$ then $x_i<n$ for all $i$ so $\tfrac{1}{x_i}>\tfrac{1}{n}$, and hence $\sum_{i=1}^n\tfrac{1}{x_i}>n\cdot\frac{1}{n}=1$. $\endgroup$
    – Servaes
    Commented Feb 14, 2021 at 14:40

1 Answer 1

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A recent master thesis, cited in https://oeis.org/A002966, briefly explores this old and hard problem.

There is no general formula.

The number of solutions are known only for $n \le 8$, and a conjecture is given for $n=9$.

It's also conjectured that the number of solutions grows double-exponentially.

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    $\begingroup$ When you say there is no general formula do you mean we don't know one or we've proved it doesn't exist? $\endgroup$
    – A-Level Student
    Commented Feb 14, 2021 at 16:53
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    $\begingroup$ Strictly speaking, the former. Actually, the phrase "general formula" is not, in itself, strictly well defined. Practically speaking, we can be quite sure that there is not a "nice" general formula. $\endgroup$
    – leonbloy
    Commented Feb 14, 2021 at 16:58
  • $\begingroup$ Ok then, thank you very much for the references and help! $\endgroup$
    – A-Level Student
    Commented Feb 14, 2021 at 17:34

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