# A Diophantine problem related to egyptian fractions

Consider the following system of equations:

$$\sum_{i=1}^{2n}a_i=0$$ $$\sum_{i=1}^{2n}\frac{1}{a_i}=0$$ Where for each $i$ $a_i$ is an odd integer and the $a_i$ are not necessarly distinct. A solution $(a_1,\dots,a_{2n})$ is trivial if (after some permutation of the coefficients) for each $i$ we have $$a_i=-a_{n+i}$$. I know that if $n>2$ there exist non trivial solutions. My questions are:

• What is the minimum number of variables for which there exist non trivial solutions ?
• Can you exhibit a minimal solution or at least a solution you think could be minimal ?
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Well, -1,3,3,5,5,-15 comes to mind. This is 2n variables for n=3. But you said you know there are non-trivial solutions for n>2. Did you mean "for every n>2" ? If so, what are you asking? Also, why not consider the case of an odd number of integers (some of which would be even)?

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Once you have one solution, you can add +1 and -1 to get one with two more variables. So with yours, we have solutions for all even n. –  Ross Millikan Nov 4 '10 at 13:53
This problem came up while I was dealing with a knot thery problem. That's why I'm not interested in further generalizations. I didn't mean for every n>2. –  Paolo Aceto Nov 4 '10 at 15:26
P.S. thanks for the answer –  Paolo Aceto Nov 4 '10 at 15:27
There is no solution for n=2. Your equations are $w+x+y+z=0$ and $\frac{1}{w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$. Take two variables of the same sign, say $x$ and $y$, and regard them as parameters. Then we have $zw\frac{x+y}{xy}+(x+y)=0$ or $zw=xy$ and $z+w=-(x+y)$. This has only the trivial solution. So the minimum n is 3.