Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c} + \frac{b}{c+a} + \frac{c} {a+b}$$

Using Nesbitt's inequality: $I \ge \frac 32$

I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c }^{3}+{c}^{2}a-2\, \left( b+c \right) \left( c+a \right) \left( a+b \right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2 }a+{c}^{2}b+abc $$ and stuck.

EDIT: Look like prove $I \le 2$ not a good thinking :P

  • $\begingroup$ Hmm, maybe try $a=b=1$ and $c=2$ or $c=3$ (I don't want to give you straight answer =p $\endgroup$ – Frudrururu May 26 '13 at 3:14
  • $\begingroup$ I don't think the OP is looking for a specific combo. I've found 3 with $a,b,c\leq 10$ myself. I think he's looking to prove that $I\leq 2$. $\endgroup$ – Michael Grant May 26 '13 at 3:14
  • $\begingroup$ I'm guessing that $a,b,c$ must be positive integers; otherwise, $a=0, b=c=1$ would work. $\endgroup$ – Adriano May 26 '13 at 3:19
  • $\begingroup$ Yes, $\mathbb{N}$ is the set of positive integers. $\endgroup$ – Michael Grant May 26 '13 at 3:20
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    $\begingroup$ Ha! I think I figured it out in the shower. Lovely things, showers are. Will write up when I get to a desk if someone else doesn't first. $\endgroup$ – Michael Grant May 26 '13 at 13:20

I find a solution to $n=4,$when

a=2332797891204725453580403814216955612718693675609518139813675622446336 8530351921955206357565424226029748329737767516130520072674084336131550 2597616224970927979227396663481447506019173462295157784788781420305046 5201815993661680059006448575315523206103260762210944137954571975497854 9786027663601160534574317253280344812956727894696796553762212813889660 9065956718516224446015577143267128739011935697434909021669583635379832 35022557869209259

b=1161640217306132458900911441651415023972393417197892812143262449233898 8034221463466278254018560734492913221738943224762433374574861704275058 0062902808034990817009121975186967451351814311101112040391014295321972 8784138582766210837461563508481437266175417187186208008663435889653439 7066554486263784443013141020886435995672339322997499528376940620045001 1919735272479457688230567501843839892799164246003766614214017398378635 0444307965016411

c=5054729227475450427274369484803239479825091305751388135572448603576037 6549781961422098862259430557133842304461180359698183208339647924784255 6816542651386138853492649101592171641096957016404851774814750638840260 3496289958758089911825477669004739864966841494437579004665357462952425 4130327474390635537868978719887059697148297723373566417781389238382736 3204638301684342182024187145267526992579708085994452308601529371953916 7125415529515145

This may be not the smallest solution to $n=4$,but I think the smallest one is also very huge.

I'd like to tell something about how I get this answer.

I estimate that there will be a solution to $n=4,$(I also tried $n=3$,but failed.)

Let $a,b,c$ be rational numbers,and $a+b+c=1,a,b,c>0,$we get $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{1-a-b}{a+b}=4.\tag{eq.1}$$ denote $\dfrac{a}{1-a}=s,\dfrac{b}{1-b}=t,$then $a=\dfrac{s}{1+s},b=\dfrac{t}{1+t},$ $eq.1$ is equivalent to $$s+t+\dfrac{1-st}{s+t+2st}=4,\tag{eq.2}$$ denote $s=x+y,t=x-y,$then$$\frac{4 x^3+3 x^2-4 x y^2+y^2+1}{2 \left(x^2+x-y^2\right)}=4$$ $$y^2=\frac{4 x^3-5 x^2-8 x+1}{4 x-9}$$ denote $x=\dfrac{9k+1}{4k},$then $$y^2=\frac{(13 k+1) \left(4 k^2+9 k+1\right)}{16 k^2}$$ denote $r=4ky,$then $$r^2=(13 k+1) \left(4 k^2+9 k+1\right)=1 + 22 k + 121 k^2 + 52 k^3$$ denote $X=52k,Y=52r,$then $$Y^2 = X^3 + 121 X^2 + 1144 X + 2704.\tag{eq.3}$$ So we just need to find a rational solution of $eq.3$ which makes $a,b,c>0.$

I find two basic solutions of $eq.3,(X,Y)=(-104,260)(52,728),$we denote $p=(-104,260),q=(52,728),$

At first,I find that $6q=O,$when I calculate $p,2p,3p,4p,\cdots$ ,I find that $21p$ makes $-52<X<-33.8564,$in fact,$X\approx -36.8062,$this is what we need,because when $-52<X<-33.8564,a,b,c$ will all be positive(it need some calculation).

Finally,we can get $a,b,c>0$ from $X,Y,$ and then multiply by the LCM of the denominators of $a,b,c,$ we are done.

You can see that I only use $p,$ but no $q$ or other points on $eq.3,$so I have to plus $21$ times by $p$,which makes the digits so huge.

  • 1
    $\begingroup$ This indicates there is probably not a nice solution. I will award the bounty to this answer because it is good evidence that it is impossible to enumerate the solutions by hand. (I was under the impression this was a contest problem.) $\endgroup$ – Potato May 30 '13 at 16:05
  • $\begingroup$ Could someone else with access to a CAS verify this is correct before I award the country? $\endgroup$ – Potato May 30 '13 at 16:06
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    $\begingroup$ @Potato: Mathematica gives me an exact 4 with these values. $\endgroup$ – celtschk May 30 '13 at 16:18
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    $\begingroup$ I'm typing how I get this solution,wait a moment please:) $\endgroup$ – lsr314 May 30 '13 at 16:28
  • $\begingroup$ @celtschk Thanks! $\endgroup$ – Potato May 30 '13 at 16:30

I found a few solutions $(a,b,c)$ of equatoin $$ \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=n \qquad\qquad (a,b,c,n\in\mathbb{N})\tag{1} $$ for even $n$.
And I have a doubt whether there are smaller solutions.


$\small a=4373612677928697257861252602371390152816537558161613618621437993378423467772036$;

$\small b=36875131794129999827197811565225474825492979968971970996283137471637224634055579$;

$\small c=154476802108746166441951315019919837485664325669565431700026634898253202035277999$.

This solution is much smaller, but is still big:
$\log_2 (a+b+c) \approx 266.723$, $\log_{10} (a+b+c) \approx 80.2916$.

Other solutions for $n=4$:




$\log_2(a+b+c)\approx 555.985$; $\log_{10}(a+b+c) \approx 167.368$.




$\log_2(a+b+c)\approx 950.321$; $\log_{10}(a+b+c)\approx 286.075$.





$\log_2 (a+b+c) \approx 443.063$; $\log_{10} (a+b+c) \approx 133.375$.





$\log_2(a+b+c) \approx 630.265$; $\log_{10}(a+b+c) \approx 189.729$.

Instead of $a,b,c \in \mathbb{N}$, we can consider positive rational numbers $p,q,r$: $$ p = \dfrac{a}{a+b+c}, \qquad q = \dfrac{b}{a+b+c}, \qquad r = \dfrac{c}{a+b+c} = 1-p-q. $$

So, we'll search rational solutions of equation $$ \dfrac{p}{1-p}+\dfrac{q}{1-q}+\dfrac{1-p-q}{2p} = n, \qquad (n\in \mathbb{N}), \tag{2} $$ like (eq.1) in Hecke's answer.

Denote $p=m-d$, $\quad$ $q=m+d$. $\quad$ $\quad$ $(2) \rightarrow (3)$: $$ \dfrac{m-d}{(1-m)-d} + \dfrac{m+d}{(1-m)+d} + \dfrac{1-2m}{2m} = n, \tag{3} $$ then $(3) \rightarrow (4)$:

$$ \Bigl((2n+6)m-1\Bigr)d^2 = (2n+6)m^3 - (4n+9)m^2+(2n+4)m-1. \tag{4} $$

Denote $m = \dfrac{x-1}{(2n+6)x}$, $d = \dfrac{y}{(2n+6)x}$. Then we get equation: $$ y^2 = 1 + (4n+6)x + (2n+3)^2x^2 + (8n+20)x^3, \qquad n \in \mathbb{N}; \tag{5} $$ $$ y^2 = \Bigl( 1 + (2n+5)x \Bigr) \Bigl(1 + (2n+1)x + 4x^2 \Bigr), \qquad n \in \mathbb{N}. $$

We need to find rational solutions $(x,y)$ of equation $(5)$, which give $ |d| <m<0.5$.

If $n=4$, then we consider rational points on elliptic curve $y^2=1 + 22 x + 121 x^2 + 52 x^3$.
Starting rational points on this EC:
points, where $x$ is $-2, 0, 1, \dfrac{-1}{4}, \dfrac{-5}{9}, \dfrac{-15}{13}, \dfrac{-2}{13}, \dfrac{-1}{13}, \dfrac{-3}{25}, \dfrac{168}{25}, \dfrac{611}{121}, \ldots$ .
We need to find/construct one with $x \in (-2.13278;-2.08649) \bigcup (-0.22120;-0.16667)$.

If $n=6$, then we consider rational points on elliptic curve $y^2=1 + 30 x + 225 x^2 + 68 x^3$,
where $x \in (-3.17116;-3.14447) \bigcup (-0.14965;-0.125)$.

If we'll find appropriate $(x,y)$, then $$ p = \dfrac{x-y-1}{(2n+6)x}, \qquad q = \dfrac{x+y-1}{(2n+6)x}, \qquad r = \dfrac{(2n+4)x+2}{(2n+6)x}. $$

It seems, that if $n$ is odd, then there are only 5 rational points on corresponding EC, and there is no integer solution $(a,b,c)$ of equation $(1)$.

In the most cases of even $n$ $(n=2,4,6,10,12,14,...)$ there are many rational points on corresponding EC.


This not a complete solution, just a 'case' of a problem.

When $\gcd(a+b,a+c,b+c)=1 \implies \text { } a,b,c \text { are not all odd or not all even}$

Your expression is


Denoting $a+b=k, b+c=l$ and $c+a=m$

$$klm |akm+blk+cml \implies a=p_1l, b=p_2m, c=p_3k$$

$a=p_1l, b=p_2m,$ and $ c=p_3k$ is not possible. So, we have no solution if we consider $\gcd(a+b,a+c,b+c)=1$.

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    $\begingroup$ When $\gcd(a+b,b+c,c+a)=\lambda>1$, then $x:=(a+b)/\lambda$, $y:=(b+c)/\lambda$, $z:=(c+a)/\lambda$ are integers. Thus $x + y - z = 2b/\lambda$ also is an integer, same for $a$ and $c$. If we write $a = \lambda a'/2$ etc, you can see that all $\lambda/2$ cancel out, and we have a new set of integers $(a',b',c')$ which gives the same number $I$, with $\gcd(a'+b',b'+c',c'+a')=2$. So it is enough to consider numbers where this gcd is 2. $\endgroup$ – celtschk May 26 '13 at 14:24
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    $\begingroup$ @celtschk: Yes, I considered gcd $2$, and feared about the larger gcd's. Now it's clear. We get no-solution in gcd $2$. That completes the proof.:) $\endgroup$ – Inceptio May 26 '13 at 14:28
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    $\begingroup$ I do not have time at the moment to write up a solution, but it's not that simple. We could have $\gcd(a+b, a+c, b+c)=1$ but $$\gcd(a+b, a+c)>1, \gcd(a+b, b+c)>1, \gcd(a+c, b+c)>1, \gcd(\gcd(a+b, a+c), \gcd(a+b, b+c), \gcd(a+c, b+c))=1$$ Similar considerations apply when the $\gcd$ is $2$. $\endgroup$ – Ivan Loh May 26 '13 at 16:02
  • $\begingroup$ I considered each pair's $\gcd$ to be $1$[To simplify the case] $\endgroup$ – Inceptio May 27 '13 at 4:48
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    $\begingroup$ @IvanLoh Could you please write up your solution? $\endgroup$ – Potato May 27 '13 at 18:36

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