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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

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I'm curious, how did you find the $a,b,c$ which you talk about at the end? – Wojowu Jan 5 at 22:47
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@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. – Michael Stoll Jan 6 at 9:00
    
I'm wondering about the solution you found for $n=4$: is there any way to compute a solution by approximation (continued fraction-style), such as approximating a real point of the elliptic curve? – Circonflexe Jan 7 at 8:32
    
(also, would you happen to have the numbers $a,b,c$ in copy-pasteable form? thanks :-D) – Circonflexe Jan 7 at 19:23
up vote 39 down vote accepted

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 $-D \equiv 1 \bmod B$, 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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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 $n = 19$. – Jeremy Rouse Jan 5 at 18:20
    
OK, thanks; I overlooked the positivity requirement. I'll look at larger $n$... – Michael Stoll Jan 5 at 18:27
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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! – René Jan 5 at 23:26
    
+1 very nice. A direct proof, compared to Allan's article mentioned below. – alex alexeq Jan 7 at 4:41

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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Cool. How did you get to consider this problem? – Michael Stoll Jan 6 at 8:57
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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 $n$. – Allan MacLeod Jan 6 at 9:35

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