Requested is a point set because the $\{a,b,c\}$ are intended to be positive integers.

It seems to me that the constraint is rather confusing to the persons answering here.

I am a naive person and use tools like Mathematica extensively.

So a simple entry point is:

It is trivial confirm there is no such solution with the constraints for positive integers if $n=1$ on the right hand side.

From that follows that the thesis is wrong. Another constraint is needed. $n>1$ and integer or $n \in \mathbb Z_{>1}$.

For $n=3$ there is an easy solution example set over at Reduce[a/(b+c)+b/(a+c)+c/(a+b)==3&&a>=0&&b>=0&&c>=0,{a,b,c},Integers].
There is no other constraint for that $n$ has to be odd.

An example set is $a=0$ and $b>0$ and $c=-b$ but that is not intended.

A valid example is $a=0$ and $b>0$ and $c= 2.61803 b$.

So it is comfortably to work through simplified variants of the problem.

For solution methodology considerations. The calculation starts with the complexes that warrant solutions and the completeness of solutions. As can be inferred on my reference page the restriction to integers is needed by hand and can be done by the reader. There is a certain hierarchy implented in the Mathematica built-in $Reduce$ to offer the nicest possible representation of the results.

What about even integers $n$?

Start with $n=2$. Look at Reduce[a/(b+c)+b/(a+c)+c/(a+b)==2&&a>=0&&b>=0&&c>=0,{a,b,c},Integers].

There are easy solutions for $a=0$ and $b>0$ and $c=b$. That are valid for the posed problem. So there is no real restriction for add integers.

The solution set $a>0$ and $b=0$ and $c=a$ shows one of the symmetries of the problem.

The two solutions sets are lines or on a line for integers. So they are degenerate with respect to the expectation that we are only able to solve this equation up to a surface arrangement for integers.

The next two solutions sets are complex solutions so we reject them due to our restriction to integer solutions.

The very same happens for $n=4$. We get only one solution line arrangement for $a=0$ and $b>0$ and $c=2.73205 b$ with the mentioned symmetry to interchange the variables. The three other solutions are either the negative one and in the complexes.

So the even parameters $n$ are not of bigger interest and can be always found on lines. Reduce[a/(b+c)+b/(a+c)+c/(a+b)==n&&a>=0&&b>=0&&c>=0&&n>0,{a,b,c,n},Integers] shows the general solution in dependence on the positive integers parameter $n$.

$a=0$ and $b>0$ and $c>0$ and $n=\frac{b}{c}+\frac{c}{b}$

and

$a>0$ and $b=0$ and $c>0$ and $n=\frac{a}{c}+\frac{c}{a}$

Symmetry $a$ can be interchanged with $b$.

$a>0$ and $b>0$ and $c\geq 0$ and $n=\frac{a^3+a^2(b+c)+a(b^2+3 b c + c^2)+b^3+b^2 c+b c^2 + c^3}{(a+b)(a+c)(b+c)}$

Mind that is not the complete solution set. But we have our constraints all at work and rejected the restriction to odd alone.

The first solutions sets are sums of the linear functions and the hyperbola considered the ratio of $b$ to $c$ or $a$ to $c$. That is rather simple.

The third solution is more complicated. It has as special case $c=0$. Which is easier considered in the posed equation than in the solution equation for $n$ and is than the same as the first solution. So by that all three variables are interchangeable as can be expected by looking at the defining equation of the posed problem.

So figuring out the integers remains a hard problem even by writing the general problem in solution form. But with the given solution we left the idea of surfaces and entered the concept of in the three dimensional integers space place special points.

Other dependencies than $n(a,b,c)$ are apparently not of interest.

We got one variable zero and the to others equal as the set of simply accessible solutions over the integers. By that $n=2$ got a very special value for solutions.

There is only need to discuss the third solution. But this is only a rewrite of the original problem simply allowing one variable being zero instead of being positive.

The most impressive step comes by the use of $FindInstance$ with the options number of instances. As shown by @allan-macleod the parameter $n=2$ is very interesting. So for $20$ instances You will get the solutions:

${{a->492,b->0,c->492},{a->0,b->49,c->49},{a->689,b->0,c->689},{a->0,b->1040,c->1040},{a->1232,b->0,c->1232},{a->778,b->0,c->778},{a->0,b->1582,c->1582},{a->1427,b->0,c->1427},{a->0,b->550,c->550},{a->1737,b->0,c->1737},{a->0,b->1767,c->1767},{a->1177,b->0,c->1177},{a->1104,b->0,c->1104},{a->0,b->1926,c->1926},{a->647,b->0,c->647},{a->0,b->604,c->604},{a->1348,b->0,c->1348},{a->0,b->391,c->391},{a->0,b->1911,c->1911},{a->0,b->978,c->978}} $

All that is remaining study the methods implemented in FindInstance. The hard part of this leap ahead is it works at present only for $n=2$. The error message given by FindInstance points into the direction that the hyperbolic nature of the identity is not targeted by the underlying methods in the manner needed.

So with increasing $n$ the character of the equation changes. Only for $n=2$ it is covered and we get plenty of nice solutions of the type already derived. A workaround for Mathematica users is changing the domain from integers to reals. This shows then solutions with two integers and one reals value for example $n=4$.

$0.381571\sqrt[3]{36a^3+153a^2b+1.73205\sqrt{-432a^6-1080a^5b+171a^4b^2+1642a^3b^3+171a^2b^4-1080ab^5-432b^6}+153ab^2+36b^3}-\frac{0.291193\left(-18a^2-33ab-18b^2\right)}{\sqrt[3]{36a^3+153a^2b+1.73205\sqrt{-432a^6-1080a^5b+171a^4b^2+1642a^3b^3+171a^2b^4-1080ab^5-432b^6}+153ab^2+36b^3}}+a+bba^2+\left(a+b+0.381571\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}-\frac{0.291193\left(-18a^2-33ba-18b^2\right)}{\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}}\right)a^2+b^2a+\left(a+b+0.381571\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}-\frac{0.291193\left(-18a^2-33ba-18b^2\right)}{\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}}\right)^2a+2b\left(a+b+0.381571\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}-\frac{0.291193\left(-18a^2-33ba-18b^2\right)}{\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}}\right)a+b\left(a+b+0.381571\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}-\frac{0.291193\left(-18a^2-33ba-18b^2\right)}{\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}}\right)^2+b^2\left(a+b+0.381571\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}-\frac{0.291193\left(-18a^2-33ba-18b^2\right)}{\sqrt[3]{36a^3+153ba^2+153b^2a+36b^3+1.73205\sqrt{-432a^6-1080ba^5+171b^2a^4+1642b^3a^3+171b^4a^2-1080b^5a-432b^6}}}\right)$

Despite this is just the first of three solution surfaces $c(a,b)$ for the
parameter $n=4$ this is very successful. For example $(17,17,128)$ is 3.99919 and $(16,18,128) is 3.99929. $c$ is rounded to integer.
At $(1500,1500,11297)$ the difference to 4 is only $\frac{1}{10000}$.$
(15000,15000,112967)$ is $4$ then exactly up to the numerical exactness by system rounding. So this is much smaller than the one given in the problem.

So this is an easy receipes calculating nice smaller number doing the required.

Since the parameter $n$ gets bigger the overall slope changes but the principal is clear. Highest order is $6$ confirming the results again from @allan-macleod for $n>2$. This is a numerical approach for the possible surfaces that describe the result without taking the restriction on integers into account first. As mentioned this is due to solvability warranty in the complexes. Get all solutions and then restrict to integers. This make it an approximation problem for integer numbers over pairs of integers.

All this solution have a small imaginary part that can be ignored and a covalue that has only to be unequal zeros, which is always true for integer pairs trivially.

The named surface is approximately since numerical constants appear depending on the parameter $n=4$. The precision can be raised but this does not improve the speed to find the third integer.

As for the case $n=2$ I expect there to be infinitely many triples matching $n=4$ in the very large value region to arbitrary exactness.

Numerics is rather different to the work of @allan-macleod. Thanks for the inspiration.

coollecture onmath&physicsentitled "From Moonshine to Black Holes: Number Theory in Mathematics and Physics", which I saw this week. The physicist Jeffrey Harvey had this problem and highlighted it on his video at 20:00 and answers it at 25:06. $\endgroup$8more comments