5
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So it turns out that there exist positive integers a, b, c and n, such that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=n.$ See Estimating the size of solutions of a diophantine equation

Now I am wondering what happens in the general case when we have m>3 variables?

$\frac{a_1}{a_2+\ldots+a_m}+\frac{a_2}{a_1+a_3+\ldots+a_m}+\ldots+\frac{a_m}{a_1+\ldots+a_{m-1}}=n.$

Interesting questions:

  1. Do solutions exist for all $m \geq 4$?

  2. Which $n$ have solutions?

So far solutions were found for all $3 \leq m \leq 30$. So far for $m \geq 4$, the following $n \in \{2,3,4,5\}$ have solutions.

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  • $\begingroup$ For $m=4$, take $(a,b,c,d)=(42,18,5,5)$. It gives $n=2$. $\endgroup$ – T. Amdeberhan Mar 11 '17 at 23:49
1
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Let's use the notation $(a_1, \ldots, a_m, n)$ to represent a solution. Then we have:

m=4: (1, 8, 66, 101, 2), (5, 5, 18, 42, 2), (5, 11, 59, 200, 3)

m=5: (1, 1, 45, 133, 261, 2) (5, 21, 35, 35, 149, 2), (22, 28, 35, 35, 188, 2)

m=6: (1, 1, 1, 12, 25, 60, 2), (3, 3, 3, 3, 58, 98, 2), (3, 3, 3, 14, 17, 62, 2), (5, 5, 18, 25, 25, 122, 2), (2, 2, 13, 13, 21, 138, 3)

m=7: (1, 1, 1, 1, 5, 17, 39, 2), (1, 1, 1, 1, 11, 11, 40, 2), (1, 1, 1, 1, 26, 32, 94, 2), (2, 2, 6, 6, 6, 13, 55, 2)

Perhaps for $m \geq 4$ there is always a solution for $n=2$ containing $(m-3)$ ones? If so, then this would form the basis of a proof that for $n=2$ there are always solutions for all $m \geq 4$.

Update I have used the above construction with $(m-3)$ ones to find new solutions. Here I will use the following format to describe a solution $(m,a_1,a_2,a_3,n)$. Here is what I found: (8, 1, 12, 27, 2), (9, 6, 14, 40, 2), (10, 1, 6, 22, 2), (11, 1, 1, 16, 2), (12, 1, 12, 34, 2), (13, 1, 85, 129, 2), (15, 1, 575, 687, 2), (16, 1, 55, 184, 3), (17, 117, 1273, 1845, 2), (18, 11, 44, 106, 2), (22, 1, 34, 82, 2), (24, 1, 8, 48, 2), (26, 1, 15, 106, 3), (27, 1, 30, 265, 5), (27, 1, 107, 187, 2), (30, 105, 222, 531, 2).

UPDATE 2 Instead of ones we can use $(m-3)$ values of $r$. This allows us to find all the missing solutions up to $m=30$. Here I will use the format $(r,m,a_1,a_2,a_3,n)$: (3, 14, 220, 395, 972, 2), (4, 19, 41, 510, 861, 2), (5, 20, 176, 936, 1715, 2), (2, 21, 203, 406, 965, 2), (2, 23, 87, 213, 512, 2), (4, 25, 89, 89, 418, 2), (3, 28, 13, 1232, 1683, 2), (3, 29, 91, 325, 741, 2).

Also I was able to find a solution with $n=4$: (1, 34, 43, 43, 442, 4).

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  • 1
    $\begingroup$ Why do you write the number? If solutions are needed, write down the parameterization of the solutions. $\endgroup$ – individ Mar 13 '17 at 7:41
  • 1
    $\begingroup$ Which number are you talking about? $\endgroup$ – Dmitry Kamenetsky Mar 13 '17 at 11:29

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