Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$

Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$.

Does it possible to find all tuples $(a,b,c,n)$ such that:

$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$

Solutions Collecting From Web of "Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$"

A linear diophantine equation
$$ax+by=c$$
where $c$ is divisible by $d=gcd(a,b)$, has solution
$$a=a_{0}+\frac{b}{d}\cdot k$$
$$b=b_{0}-\frac{a}{d}\cdot k$$
However, your equation is not linear, since
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=n$$
$$\frac{\left(a+b\right)ab+\left(b+c\right)bc+\left(a+c\right)ac}{abc}=n$$
Note here you are interested in combinations of $\left(a,b,c\right)$ such that this quotient is an integer $n$. Also,
$$a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}=abcn$$
proves it’s not possible to represent $\left(a,b,c\right)$ in the same terms of $k$ as a linear diophantine equation, so they don’t share a straight-forward relationship, but this last equation is very interesting, since it is contained in the cubic expansion $\left(a+b+c\right)^{3}$.

Then,
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(abcn\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3abc\left(n+2\right)$$
$$\left(a+b+c\right)^{3}-\left(a^{3}+b^{3}+c^{3}\right)=3abc\left(n+2\right)$$
As you can see, this difference is divisible by 3, so using congruence is an excellent idea:
$$\left(a+b+c\right)^{3}\equiv a^{3}+b^{3}+c^{3}\;\left(mod\;\;3\right)$$
So, given that $gcd\left(a,b,c\right)=1$, every combination of $\left(a,b,c\right)$ where the cube of the sum is congruent to the sum of the cubes $\left(mod\;\;3\right)$ is a possible answer.

For instance, $\left(1,1066,3977\right)$ and $\left(1,1598,4182\right)$ are valid answers, but it’s difficult to trace an easy linear relation between $1$ and the other values.

First note that we can assume $a,b,c \in \mathbb{Q}$, without loss of generality. A rational solution can then be scaled to integers.

Combining into a single fraction gives the quadratic in $a$
\begin{equation*}
a^2+\frac{b^2-nbc+c^2}{b+c}a+bc=0
\end{equation*}
and, for rational solutions, the discriminant must be a rational square.

Thus, the quartic
\begin{equation*}
D^2=b^4-2(n+2)b^3c+(n^2-6)b^2c^2-2(n+2)bc^3+c^4
\end{equation*}
must have rational solutions.

This quartic is birationally equivalent to the elliptic curve
\begin{equation*}
V^2=U^3+(n^2-12)U^2+16(n+3)U
\end{equation*}
with
\begin{equation*}
\frac{b}{c}=\frac{V+(n+2)U}{2(U-4(n+3))}
\end{equation*}

The elliptic curve is singular when $n=-2,-3,6$. $n=6$, for example, corresponds to $a=b=c=K$ as a solution.

If $n \ne -2,-3,6$, the curve has $5$ finite torsion points at $(0,0)$, $(4,\pm 4(n+2))$ and $(4n+12,\pm 4(n+2)(n+3))$ none of which give a solution. For $n=7$, there are a further $6$ torsion points which lead to the solutions $(a,b,c)=(1,1,2)$ and $(a,b,c)=(1,2,2)$.

Thus, if $n \ne -3,-2,6,7$, we need the elliptic curve to have rank greater than zero to find a solution. Computations using the Birch and Swinnerton-Dyer conjecture suggest the first few solutions are with $n=8,11,12,15,\ldots$.

For example, the $n=15$ curve has a generator $(-36,468)$ which gives the solution $a=2, b=3, c=15$. As $n$ gets larger, the size of solutions generally increases.

Add +3 both sides
We get $(a+b+c)(\frac{ab+bc+ca}{abc})=n+3$
Since $gcd(ab+bc+ca,abc)=1$ beacuse $(a,b,c)=1$ we should have $(a+b+c)\vdots (abc)$ and then we can get solutions $(1,1,1),(1,1,2),(1,2,3)$

I’ll describe one approach which refers toElliptic Curves.

First, denote $$p=\dfrac{a}{a+b+c}, \quad q=\dfrac{b}{a+b+c}, \quad r = \dfrac{c}{a+b+c}=1-p-q.\tag{1}$$

Then after adding $\dfrac{c}{c}+\dfrac{a}{a}+\dfrac{b}{b}=3$ to starting equation, we’ll get
$$
\dfrac{1}{p}+\dfrac{1}{q} + \dfrac{1}{r} = n+3;\tag{2}
$$
Then denote $$p=m-d,\quad q=m+d, \; \mbox{ where } |d|<m<1/2,$$
and $(2)\Rightarrow (3)$:
$$
\dfrac{1}{m-d}+\dfrac{1}{m+d}+\dfrac{1}{1-2m} = n+3;\tag{3}
$$
$$
\dfrac{2m(1-2m)+(m^2-d^2)}{(m^2-d^2)(1-2m)} = n+3;
$$
after some transformations we’ll get equation for $(m,d)$:
$$
(2(n+3)m-n-2)d^2 = 2(n+3)m^3-(n+6)m^2+2m.
$$

Finally, denote
$$
m = \dfrac{(n+2)x+1}{2(n+3)x}, \quad d = \dfrac{y}{2(n+3)x};
$$
after some simplifying we’ll get equation for $(x,y)$:
$$
y^2 = 4(n+2)x^3 + n^2x^2+2nx+1.\tag{4}
$$

Equation $(4)$ describes certain Elliptic Curve (EC).

Now,

  • try to find a few starting rational points of EC (using brute force search, for example);
  • applying the group law, find other (as much as possible) rational points of EC;
  • finally, select such points only, for which $|m|<d<1/2$.

This method (and the structure of equation) is closely related to method of finding similar problem solutions.

Since we’ll obtain huge integer numbers $a,b,c$ after few steps of walking along EC, I’ll show reasonably large ones only:

\begin{array}{|c|c|}
\hline
n & a,b,c \\
\hline
6 & 1,1,1 \\
\hline
7 & 1,1,2 \\
& 1,2,2 \\
\hline
8 & 1,2,3 \\
& 2,3,6 \\
& 8177, 8874, 22243 \\
& 11322, 28379, 30798 \\
&
264668812593,
596917970819,
765759090202 \\
&
579432117963,
743327279754,
1676455216382 \\
&
76958469196744137591058,
134451734421662586058497,
233763322749706719184931 \\
&
140374393770070337430066,
244060702212707093922518,
426390819089416636962387 \\
& \ldots \\
\hline
11 & 2,3,10 \\
& 3,10,15 \\
& 129609938, 398619322, 665659395 \\
& 304406698, 508332555, 1563392295 \\
&
849823960453237978650627,
3053593877043831122496050,
4107252165654093769719898 \\
&
2249722029555895456095675,
3026000198480402499943863,
10873046781459417805417450 \\
& \ldots \\
\hline
12 & 1,2,6 \\
& 1,3,6 \\
& 201134, 841113, 1072474 \\
& 333459, 425182, 1778049 \\
&
13609929187686894,
57640720247049619,
71996401755236577 \\
&
17668312802572318,
22068685842562794,
93465214207141769 \\
& \ldots \\
\hline
15 & 2,3,15 \\
& 2,10,15 \\
& 952712343, 3044622267, 7974923330 \\
& 1475561727, 3865008730, 12351568370 \\
& \ldots \\
\hline
16 & ? \\
\hline
19 & ? \\
\hline
20 & ? \\
\hline
23 & 1,6,14 \\
& 3,7,42 \\
& 1621859337, 2014655278, 18651341318 \\
& 1975996071, 18293440851, 22723904794 \\
& \ldots \\
\hline
24 & ? \\
\hline
27 & 10, 77, 165 \\
& 14, 30, 231 \\
&
4430547232529081454,
36139386493080531486,
71649568303086464735 \\
&
7082090148273898842,
14040877586688470045,
114529577313119362405 \\
& \ldots \\
\hline
28 & ? \\
\hline
31 & 1,6,21 \\
& 2,7,42 \\
& 8878833, 17378998, 166742613 \\
& 9449986, 90667791, 177468746 \\
&
200321970145448798577,
2788507822582499346582,
2992032039624422248117 \\
&
747195060205269358914,
801730424379013131359,
11160191557421428133194 \\
& \ldots \\
\hline
32 & 3, 10, 65 \\
& 6, 39, 130 \\
&
1821365467193555,
20672336467139058,
33287283401295745 \\
&
7943231969284014,
12790456178192335,
145170542895601626 \\
& \ldots \\
\hline
35 & 90, 391, 2210 \\
& 207, 1170, 5083 \\
&
182127607562055810474624817130,
1320995230654532333162659755111,
4363962685714983855023196247570 \\
&
568734025435763596856102290167,
1878836507129847550250015042290,
13627445604327101620082253820563 \\
& \ldots \\
\hline
36 & ? \\
\hline
39 & 561, 6450, 13889 \\
& 1650, 3553, 40850 \\
&
19918302114889088017632015217487883650,
60569462602897456667888129359842887481,
531797592631412085961939529793102092169 \\
&
26353710650431538163999207644180706050,
231384583559667745287470357470008076450,
703616191780126367349950469574356131913 \\
& \ldots
\end{array}

To check them, one can input in WolframAlpha text like this:

(a+b)/c+(a+c)/b+(b+c)/a where (a,b,c)=(304406698, 508332555, 1563392295)

Note 1. After $n=6$, I didn’t find any $n$ of the form $n=4k+1$, $n=4k+2$, for which starting equation has solution(s).

Note 2. One can see that solutions are grouped by pairs:
$$(a_1,b_1,c_1,n)\\(a_2,b_2,c_2,n).$$
And one can see that for such pairs (if denote $M=LCM(a_1,b_1,c_1)$):
$$a_1c_2=b_1b_2 = c_1a_2 = LCM(a_1,b_1,c_1) = LCM(a_2,b_2,c_2) =M.$$
Indeed, if $(a_1,b_1,c_1,n)$ is solution of starting equation, then
$$\dfrac{a_1+b_1+c_1}{a_1}+\dfrac{a_1+b_1+c_1}{b_1}+\dfrac{a_1+b_1+c_1}{c_1}=n+3,$$
$$(a_1+b_1+c_1)\left(\dfrac{1}{a_1}+\dfrac{1}{b_1}+\dfrac{1}{c_1}\right)=n+3,$$
$$(a_1+b_1+c_1)\left(\dfrac{M/a_1}{M}+\dfrac{M/b_1}{M}+\dfrac{M/c_1}{M}\right)=n+3,$$
$$(a_1+b_1+c_1)\dfrac{\left(M/a_1+M/b_1+M/c_1\right)}{M}=n+3,$$
$$(M/a_1+M/b_1+M/c_1)\left(\dfrac{a_1+b_1+c_1}{M}\right)=n+3,$$
hence $\left(M/a_1,M/b_1,M/c_1,n\right)$ is solution too.