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Solve for $x,y,z\in\mathbb{N}$

$\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

I tried by some general methods but they didn't help me.

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Wlog. $x\le y\le z$. Then there are only finitely many cases for $x$ to consider as $\frac3x\ge \frac4{13}>\frac1x$ implies $\frac{13}4<x\le\frac{39}{4}$, i.e. $4\le x\le 9$. For each of these $x$, there are again only finitely many $y$ to consider, as we have $\frac2y\ge\frac4{13}-\frac1x>\frac1y$, i.e. $\frac1{\frac4{13}-\frac1x}<y\le \frac2{\frac4{13}-\frac1x}$. For each such $y$ check if $z=\frac1{\frac4{13}-\frac1x-\frac1y}$ is an interger (and $x\le y\le z$). So all in all this is a finite task.

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I am going to try brute force. The smallest $x$ for which $\frac1x < \frac{4}{13}$ is $x=4$, so I will try that, and then I need $$\frac1y + \frac1z = \frac4{13}-\frac14 = \frac3{52}.$$

Now it is clear that $\frac3{52} = \frac1{52} + \frac2{52} = \frac1{52} +\frac1{26}$ so I am done. But if I didn't have that happy inspiration, I could repeat the process and try $y=18$ because that is the smallest $y$ for which $\frac1y \le \frac3{52}$. Then I find $$\frac1z = \frac3{52} - \frac1{18} = \frac1{468}$$ and I have a second solution.

I cannot claim this is a general method, but it did work with minimal effort and no theory.

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  • $\begingroup$ The greedy algorithm is guaranteed to work, but it can be very inefficient in the sense that there may be shorter expansions. But it’s surely the obvious thing to try here if one doesn’t have any cleverer ideas. $\endgroup$
    – Brian M. Scott
    Nov 10, 2014 at 18:52
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    $\begingroup$ It is guaranteed to find an expansion, but it may not find the shortest one. This problem required an expansion of three terms. The greedy algorithm found one, in this case, but is not guaranteed to do so. $\endgroup$
    – MJD
    Nov 10, 2014 at 18:55
  • $\begingroup$ I know: that’s pretty much what I said in my comment. $\endgroup$
    – Brian M. Scott
    Nov 10, 2014 at 18:57
  • $\begingroup$ You said it was guaranteed to work, but for this problem, that is not the case. $\endgroup$
    – MJD
    Nov 10, 2014 at 18:58
  • $\begingroup$ It is guaranteed to work in the sense in which I meant it: it always produces an Egyptian fraction expansion. The ‘but it’ clause clearly implies that it might not have produced a $3$-term expansion here, and the final sentence clearly implies that it might not have worked. $\endgroup$
    – Brian M. Scott
    Nov 10, 2014 at 18:59
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Hint: Erdős-Strauss-Conjecture and Egyptian-Fraction-Expansion

You can get

$$\frac 4 {13} = \frac 1 4+ \frac 1 {18}+ \frac 1 {468}$$

Try using the expansion algorithm:

$$\frac{x}{y}=\frac{1}{\lceil y/x\rceil}+\underbrace{\frac{(-y)\,\bmod\, x}{y\lceil y/x\rceil}}_{*}$$

On the LHS you have your input. Then you can calculate the two terms on the RHS. The first one is the first fraction of the expansion while you have to expand the second one $(*)$ again with the same algorithm.

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  • $\begingroup$ But I want solution to this case. Can you state some method to find it? $\endgroup$
    – Satvik Mashkaria
    Nov 10, 2014 at 18:31
  • $\begingroup$ I did post the solution and you can use the given expansion algorithm. $\endgroup$
    – flawr
    Nov 10, 2014 at 18:32
  • $\begingroup$ Is there a condition on $m,n$ for which $\frac{m}{n}=x^-1+y^-1$ is solvable where $x,y\in\mathbb{N}$? $\endgroup$
    – Satvik Mashkaria
    Nov 10, 2014 at 18:35
  • $\begingroup$ Hint: Write $m/n$ as $(x+y)/(xy)$ but this is a totally different problem. $\endgroup$
    – flawr
    Nov 10, 2014 at 18:35
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    $\begingroup$ Egyptian Fractions. I learnt something new today. It's interesting how the greedy algorithm leads to inelegant results. $\endgroup$
    – Simon S
    Nov 10, 2014 at 18:39

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