A. Is there natural numbers $a,b,c$ such that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an odd natural number ?

(I do not know any such numbers).

B. Suppose that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an even natural number ($a,b,c $ are still natural numbers) then is there any way to estimate the minimum of $a,b $ and $c$ ?

The smallest solution that I know for $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4 $ is: enter image description here

In text (from the comment below of @djsadinoff): $$\scriptsize{a = 4373612677928697257861252602371390152816537558161613618621437993378423467772036}$$ $$\scriptsize{b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579}$$ $$\scriptsize{c = 154476802108746166441951315019919837485664325669565431700026634898253202035277999} $$

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    $\begingroup$ I'm curious, how did you find the $a,b,c$ which you talk about at the end? $\endgroup$
    – Wojowu
    Jan 5, 2016 at 22:47
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    $\begingroup$ @Wojowu : I don'tknow how alex alexeq found them, but what you can do is the following. You use a computer algebra system like Magma or Sage to find the generators of the group $E'_4({\mathbb Q})$ (compare my answer), then you enumerate the points on the non-identity component by increasing height and check if they give positive $a,b,c$. The numbers $a,b,c$ given in the question come from the first set of six points one finds in this way. $\endgroup$ Jan 6, 2016 at 9:00
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    $\begingroup$ a = 4373612677928697257861252602371390152816537558161613618621437993378423467772036; b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579; c = 154476802108746166441951315019919837485664325669565431700026634898253202035277999; $\endgroup$
    – djsadinoff
    Mar 9, 2017 at 11:57
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    $\begingroup$ There is a cool lecture on math & physics entitled "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. $\endgroup$ Sep 8, 2017 at 22:56
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    $\begingroup$ Just adding link to Alon Amit's quora answer which which references this MO question and is a more elementary overview $\endgroup$
    – Dabed
    Aug 8, 2021 at 13:48

3 Answers 3


This problem turned out to be much more interesting than I originally thought. Let me give my solution, which seems to be slightly different from (but essentially the same as) the solution in the paper by Bremner and MacLeod (see Allan MacLeod's answer).

Theorem. Let $a,b,c$ be positive integers. Then $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can never be an odd integer.

Let $n$ be a positive odd integer. The equation $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$ implies $$a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$ This describes a smooth cubic curve $E_n$ in the projective plane that has at least six rational points (of the form $(1:-1:0)$ and $(1:-1:1)$ and their cyclic permutations). Declaring one of these to be the origin, $E_n$ is an elliptic curve over $\mathbb Q$. Bringing $E_n$ in Weierstrass form, we obtain the isomorphic curve $$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr) =: x(x^2 + Ax + B).$$ If $n = 1$, then there are obviously no positive solutions, so we assume $n \ge 3$. Then $E_n(\mathbb R)$ has two connected components, one of which contains the six `trivial' points but no points with positive coordinates, whereas the other component does contain positive points. In the model $E'_n$, this component consists of points with negative $x$-coordinate.

Claim. If $(\xi,\eta) \in E'_n(\mathbb Q)$, then $\xi \ge 0$.

This clearly implies the statement of the theorem.

To show the claim, let $D = 2n + 5$. Then $D$ is odd, positive, coprime with $B$ and divides $A^2 - 4B = (2n-3)(2n+5)^3$. If $p$ is an odd prime dividing $B$, then $n \equiv -3 \bmod p$ and so $-D \equiv 1 \bmod p$. The equation $B x^2 - D y^2 = z^2$ has the solution $(x,y,z)=(1,4,4)$, so the Hilbert symbol $(B, -D)_p = 1$ for all primes $p$. We will show:

If $(\xi,\eta) \in E'_n(\mathbb Q)$ with $\xi \neq 0$, then $(\xi, -D)_p = 1$ for all primes $p$.

Given this, the product formula for the Hilbert symbol implies $(\xi, -D)_\infty = 1$ and so $\xi > 0$ (since $-D < 0$).

Note that $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p$. We first consider odd $p$. We note that when $\xi$ is not a $p$-adic integer, then $\xi$ must be a $p$-adic square, so $(\xi, -D)_p = 1$. So we can assume that $\xi \in {\mathbb Z}_p$. There are three cases.

  1. $p$ divides neither $B$ nor $D$. If $\xi \in {\mathbb Z}_p^\times$, then $(\xi, -D)_p = 1$, since both entries are $p$-adic units. Otherwise, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.
  2. $p$ divides $B$. Then $-D \equiv 1 \bmod p$, so $-D$ is a $p$-adic square, hence $(\xi, -D)_p = 1$.
  3. $p$ divides $D$. Then $x^2 + Ax + B \equiv (x + A/2)^2 \bmod p$. So if $\xi \in {\mathbb Z}_p^\times$, then $\xi$ must be a square mod $p$, and $(\xi, -D)_p = 1$. If $\xi$ is divisible by $p$, then as before, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.

It remains to consider $p = 2$. If $n \equiv 1 \bmod 4$, then $-D \equiv 1 \bmod 8$, so $(\xi, -D)_2 = 1$ for all $\xi$. If $n \equiv 3 \bmod 4$, then $-D \equiv 5 \bmod 8$, so $(\xi, -D)_2 = (-1)^{v_2(\xi)}$, and we have to show that the 2-adic valuation of $\xi$ must be even. Note that in this case $v_2(B) = 6$ and $A \equiv -3 \bmod 8$. If $v_2(\xi)$ is odd, then exactly one of the three terms $\xi^3$, $A \xi^2$, $B \xi$ has minimal 2-adic valuation, which must be even, so it cannot be the first or the third term. This reduces us to $\nu := v_2(\xi) \in \{1,3,5\}$. One then easily checks that $\xi(\xi^2 + A\xi + B) = 4^\nu u$ with $u \equiv -1 \bmod 4$ when $\nu = 1$ or $5$ and $u \equiv -3 \bmod 8$ when $\nu = 3$. In all cases, $u$ cannot be a square, and so points with $x$-coordinate $\xi$ cannot exist. This concludes the proof.

Note that when $n$ is even, we have $-D \equiv 3 \bmod 4$ and also $v_2(B) = 5$, so we lose control over the 2-adic Hilbert symbol.

This is the previous version of this answer, which I leave here, since it may contain some points of interest.

The equation $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$ gives rise to the elliptic curve $$E_n \colon a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$ You are asking for rational points on this curve (such that $a+b, b+c, c+a \neq 0$). For odd positive $n$ up to and including 17, this is a curve of rank zero (with 6 rational points), whereas for $n = 19$, it has rank 1. Therefore $E_{19}$ has infinitely many rational points, and your equation has infinitely many solutions for $n = 19$. I'll do the computations and find one explicitly.

EDIT: As pointed out by Jeremy Rouse in a comment below, the integral solutions for $n = 19$ are not positive. More precisely, the real points $E_n(\mathbb R)$ form two connected components (the discriminant of $E_n$ is positive), and it is the non-identity component that contains points with all positive coordinates (taking as the identity one of the six points like $(1:-1:0)$ or $(1:1:-1)$). So the question is whether there is odd $n$ such that there is a rational point on the non-identity component; then the rational points will be dense on this component and so there will be positive solutions. So far, no such $n$ turned up, even though there are many such that $E_n$ has positive rank.

FURTHER EDIT: I suspect that there really is no odd $n > 0$ such that $E_n$ has rational points on the non-identity component. One way of checking this for any given $n$ is to do (half of) a 2-isogeny descent on $E_n$. This produces a number of curves of the form $C_u \colon y^2 = u x^4 + v x^2 + w$ where $v = 4n(n+3)-3$ and $uw = 32(n+3)$ that are unramified double covers of $E_n$. We consider the curves $C_u$ that have points over all completions of $\mathbb Q$. Then every rational point on $E_n$ is the image of a rational point on one of these curves $C_u$. Doing the computation, one obtains a set of curves $C_u$ that all have $u > 0$ (this is only experimental; I checked it for $n$ up to 9999). But if $u > 0$, then [$C_u$ has only one real component — this is wrong, but the following is OK] the image of $C_u(\mathbb R)$ in $E_n(\mathbb R)$ is the identity component, so there can be no rational point on the other component. My feeling is that there might be a Brauer-Manin obstruction to the existence of rational points on the non-identity component for odd $n$, but I don't have enough time to check this. A possible approach would be to note that $$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr)$$ is isomorphic to $E_n$. If we can find a positive integer $d(n)$ such that for all rational points $(\xi,\eta) \in E'_n(\mathbb Q)$ (with $\xi \neq 0$) the product $\prod_p (\xi, -d(n))_p$ of Hilbert symbols (over all finite places) is always $+1$, then the claim follows from the product formula for the Hilbert symbol and $(\xi, -d(n))_\infty = -1$ for $\xi < 0$.

SUCCESS: For odd $n \ge 3$, $d(n) = 2n-3$ works. One can check that $(\xi, 3-2n)_p = 1$ for all primes $p$. Details later (it is getting late). Actually, $d(n) = -5-2n$ works better. See above.

Note that for even $n$, there usually are $C_u$ with $u < 0$ when $E_n$ has positive rank (the first exception seems to be $n = 40$). So I would expect the Brauer-Manin obstruction to result from an interaction between $p = 2$ and the infinite place.

For $n = 4$, the curve has also rank 1, which explains the existence of solutions. I'll try to check if there are smaller ones than that given by you.

EDIT: The given solution is really the smallest (positive) one. The next larger one has numbers of 167 to 168 digits.

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    $\begingroup$ The OP is requesting $a$, $b$ and $c$ to be positive, which appears to correspond to points on the non-identity component of the Weierstrass model of $E_{n}$. I don't think there are such points for $n = 19$. $\endgroup$ Jan 5, 2016 at 18:20
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    $\begingroup$ OK, thanks; I overlooked the positivity requirement. I'll look at larger $n$... $\endgroup$ Jan 5, 2016 at 18:27
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    $\begingroup$ This must be one of the easiest-to-state (and most attractive!) problems in number theory whose solution "requires" the Brauer-Manin obstruction. Very cool! $\endgroup$
    – R.P.
    Jan 5, 2016 at 23:26
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    $\begingroup$ @GHfromMO You should actually stop reading at the first horizontal line; what is below are my somewhat incremental thoughts leading to the solution that is spelled out in some detail above the line. To get the isomorphism, one changes coordinates so that one of the inflection points is at infinity and the tangent at that points is the line at infinity (this works here, since there is a rational inflection point; in general, there is still an isomorphism with a curve in Weierstrass form, but it is more complicated)... $\endgroup$ Mar 25, 2017 at 18:11
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    $\begingroup$ @GHfromMO (continued) In the concrete case, an isomorphism is given by setting $x = 4(n+3)(a+b+2c)/(c-(n+2)(a+b))$ and $y = (8n^2+44n+60)(a-b)/(c-(n+2)(a+b))$ (according to Magma). $\endgroup$ Mar 25, 2017 at 18:14

This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et Informaticae, pages 29-41.

It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can be of truly enormous size - much larger than the example given.

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    $\begingroup$ Cool. How did you get to consider this problem? $\endgroup$ Jan 6, 2016 at 8:57
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    $\begingroup$ I was playing around with several cubic representation problems in the style of previous work by Andrew and Richard Guy. The numerical results were fascinating so I sent the initial work to Andrew, who very quickly proved the result about odd $n$. $\endgroup$ Jan 6, 2016 at 9:35
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    $\begingroup$ Adding a link to a YouTube video that discusses the case $n = 4$: youtube.com/watch?v=Ct3lCfgJV_A $\endgroup$
    – KConrad
    Dec 19, 2021 at 4:18

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


$a>0$ 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 $n=4$.


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.


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