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From my friend, he gives me a competition question:

"How many solution $(a,b,c,d)$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ have where $a,b,c,d$ are positive integers? (the size of $a,b,c,d$ doesn't matter, either one can be the biggest or smallest, and they are not necessarily distinct)"

I want to ask if there is any solution shorter than mine? I think mine is too long, and maybe yields a wrong answer.


My solution: WLOG, let $a\leq b\leq c\leq d$ $$1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq \frac{4}{a}$$ $$a\leq4$$ Because $a=1$ yields no solution, so consider $a=2,3,4$

Case 1:$a=2$, then $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$
Do that again: $\frac{1}{2}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq\frac{3}{b}$, so $b\leq 6$.
Let $b=6$, then $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$, going to $$(c-3)(d-3)=9$$ $$(c,d)=(4,12),(6,6)$$ so in the case have: $(a,b,c,d)=(2,6,4,12),(2,6,6,6)$ then eliminate some case not satisfy $a\leq b\leq c\leq d$
Then going through when $b=5$,$b=4$,$b=3$... yields $184$ distinct solutions.

Case 2: Following the same procedure as Case 1... yields $18$ solutions.

Case 3: As above... yields only a solution which is $(4,4,4,4)$

Conclude it, the equation has $203$ solutions.


That is my solution, I wrote it using one and a half piece of A4 paper, I have recently tried $abcd=abc+abd+acd+bcd$ but don't know how to continue, or should I use Vieta theorem?
---After first edit---
According to Robert Z, I had miscount quadruplet $(3,4,4,6)$ which add up the count to $215$ solutions.
-- After last edit --
Seems like there is no faster solution, I will close this question and marked as solved. Thanks to everyone who spend effort to my question.

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  • $\begingroup$ I’m sure I’ve see something similar on this site with five fractions that got many upvotes and much attention. Unfortunately I cannot find it. $\endgroup$
    – Szeto
    Apr 21, 2018 at 5:14
  • $\begingroup$ Okay... I will try to find it. $\endgroup$ Apr 21, 2018 at 5:19
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    $\begingroup$ See also A002966 Egyptian fractions: number of solutions of .... $\endgroup$
    – dxiv
    Apr 21, 2018 at 5:24
  • $\begingroup$ In the case if $\min\{a,b,c,d\}=3$, we have $30$ (not $18$) solutions. $$ 3, 3 ,4 ,12 \;(12 \;solutions) \\ 3, 4, 4, 6\; (12 \;solutions) \\ 3, 3, 6, 6\; (6\; solutions) $$ So totally $215$ solutions. $\endgroup$
    – Oleg567
    Apr 21, 2018 at 5:34
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    $\begingroup$ @kelvinhong方 You just sort those by the number of distinct values and multiply by the appropriate factor. For example, they list $6$ solutions where all values are distinct, so those would count as $6 \cdot 4!$ for you. $\endgroup$
    – dxiv
    Apr 21, 2018 at 5:49

3 Answers 3

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In my opinion your approach is fine and I am not aware of a faster method.

However I got a different number of solutions. Assuming that $a\leq b\leq c\leq d$ then

1) if $a=2$ and $b=3$ then $(c,d)$ can be $$(7,42),\;(8,24),\;(9,18),\;(10,15),\;(12,12).$$

2) if $a=2$ and $b=4$ then $(c,d)$ can be $$(5,20),\;(6,12),\;(8,8).$$

3) if $a=2$ and $b\geq 5$ then $(b,c,d)$ can be $$(5,5,10),\;(6,6,6).$$

4) if $a=3$ then $(b,c,d)$ can be $$(3,4,12),\;(3,6,6),\;(4,4,6).$$

5) if $a=4$ then $(b,c,d)=(4,4,4)$.

Hence rearranging the $14$ ordered solutions we find the total number of solutions: $$6\cdot 4!+5\cdot \frac{4!}{2}+1\cdot \frac{4!}{2!2!}+1\cdot \frac{4!}{3!}+1=215.$$

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    $\begingroup$ Maybe not count in $(2,5,5,10),(2,6,6,6)$? And I had miscount $(3,4,4,6)$ So my new answer is $215$ of them. $\endgroup$ Apr 21, 2018 at 5:29
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    $\begingroup$ @kelvinhong方 Yes, I got them on paper but I forgot to write them down. $\endgroup$
    – Robert Z
    Apr 21, 2018 at 5:41
  • $\begingroup$ Okay, but your solution seems same to mine. Isn't there faster way to do this? Other than restrict the range of $a,b,c,d$ and using permutation. $\endgroup$ Apr 21, 2018 at 5:43
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    $\begingroup$ I am not aware of a faster method. I just pointed out that your enumeration was not correct. $\endgroup$
    – Robert Z
    Apr 21, 2018 at 5:47
  • $\begingroup$ Okay, but also thanks to your clarification! $\endgroup$ Apr 21, 2018 at 5:51
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Above equation shown below has parametric form:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$

$a=k(3-k)/2k$

$b=k(3-k)/k$

$a=3(k-3)/2k$

$a=3(k-3)/k$

For $k=-7$ we get $(a,b,c,d)=[5,10,(15/7),(30/7)]$

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  • $\begingroup$ I cannot understand your solution, and it also didn't vendor correct answer. $\endgroup$ Apr 23, 2018 at 3:07
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There was a typo in my previous answer yesterday for the below equation:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$

The variables (a,b,a,a) should read as (a,b,c,d) & is shown below:

$a=k(3-k)/2k$

$b=k(3-k)/k$

$c=3(k-3)/2k$

$d=3(k-3)/k$

Hence for $k=-7$ , we get:

$(1/5)+(1/10)+(7/15)+(7/30)=1$

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  • $\begingroup$ Oh I see, but the question states that $a,b,c,d$ must be positive integer hence not $\frac{15}{7}$ or $\frac{30}{7}$. $\endgroup$ Apr 24, 2018 at 3:13

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