39

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

zyx
  • 36,077
Xeing
  • 3,017
  • Hmm, maybe try $a=b=1$ and $c=2$ or $c=3$ (I don't want to give you straight answer =p –  May 26 '13 at 03:14
  • 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 $I\leq 2$. – Michael Grant May 26 '13 at 03:14
  • I'm guessing that $a,b,c$ must be positive integers; otherwise, $a=0, b=c=1$ would work. – Adriano May 26 '13 at 03:19
  • Yes, $\mathbb{N}$ is the set of positive integers. – Michael Grant May 26 '13 at 03:20
  • Where is this problem from? – Potato May 26 '13 at 03:39
  • I don't know why you are trying to prove that, when $c=3(a+b)$ you have $I=\frac{a}{b+c}
    • \frac{b}{c+a} + \frac{c} {a+b} \ge 3$.
     – Inceptio May 26 '13 at 12:48
  • But $I$ is not necessarily an integer in that case. – Michael Grant May 26 '13 at 12:57
  • @MichaelC.Grant: Yeah. But trying to prove the inequality is a bad way to go. – Inceptio May 26 '13 at 13:00
  • Agreed. My interpretation of the question is this: $I\in\mathbb{N}\Longrightarrow I\leq 2$; alternatively, $I>2\Longrightarrow I\not\in\mathbb{N}$. @D3roX4, can you please confirm this interpretation? It is clear from the comments, including those that I have seen that have since been deleted, that this question could use some clarification. It is also clear there is a lot of interest in this question! – Michael Grant May 26 '13 at 13:02
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    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. – Michael Grant May 26 '13 at 13:20
  • @MichaelC.Grant: So, it does or doesn't have solution? – Inceptio May 26 '13 at 13:37
  • I believe I have a proof to my interpretation of the question: that $I>2\Longrightarrow I\not\in\mathbb{Z}$. I am working out a couple of kinks. – Michael Grant May 26 '13 at 14:23
  • The kinks remain. Unless I can work them out I will post what I have tonight and let someone else fix it. I'm pretty sure it's close. – Michael Grant May 26 '13 at 21:04
  • Well, drat. Actually it's worse than I thought. My work will not be helpful to others. Sorry. Carry on – Michael Grant May 27 '13 at 03:18
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    See also the discussion at https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation – Gerry Myerson Jan 29 '23 at 22:32

3 Answers3

34

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.

lsr314
  • 16,048
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    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.) – Potato May 30 '13 at 16:05
  • Could someone else with access to a CAS verify this is correct before I award the country? – Potato May 30 '13 at 16:06
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    @Potato: Mathematica gives me an exact 4 with these values. – celtschk May 30 '13 at 16:18
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    I'm typing how I get this solution,wait a moment please:) – lsr314 May 30 '13 at 16:28
  • @celtschk Thanks! – Potato May 30 '13 at 16:30
  • Are there smaller solutions with $a,b,c > 0$ in the set $p, p+q, p+2q, p+3q, p+4q, p+5q$? – zyx May 30 '13 at 17:17
  • @zyx,I tried many small $up+vq$,no one makes $a,b,c>0,$but I didn't try all the conditions for $u<20,v<6.$ – lsr314 May 30 '13 at 17:22
  • It could be that the set $a,b,c > 0$, when pushed through all the changes of variable, occupies a smaller fraction of the volume in the invariant measure on the curve than the 1/4 of $(a,b,c)/{\mathbb{R}^*}$ space, so one has to look further. Or $p$ is small in Archimedean norm (compared to the height of the curve) and far from the $(a,b,c)>0$ locus. – zyx May 30 '13 at 17:27
  • What is Rank and Generator of an Elliptic curve ? – Guruprasad Feb 23 '25 at 21:53
20

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.

$\color{#FF4400}{n=4:}$

$\small a=4373612677928697257861252602371390152816537558161613618621437993378423467772036$;

$\small b=36875131794129999827197811565225474825492979968971970996283137471637224634055579$;

$\small c=154476802108746166441951315019919837485664325669565431700026634898253202035277999$.

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 $n=4$:

a=16666476865438449865846131095313531540647604679654766832109616387367203990642764342248100534807579493874453954854925352739900051220936419971671875594417036870073291371;

b=184386514670723295219914666691038096275031765336404340516686430257803895506237580602582859039981257570380161221662398153794290821569045182385603418867509209632768359835;

c=32343421153825592353880655285224263330451946573450847101645239147091638517651250940206853612606768544181415355352136077327300271806129063833025389772729796460799697289;

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

a=1054210182683112310528012408530531909717229064191793536540847847817849001214642792626066010344383473173101972948978951703027097154519698536728956323881063669558925110120619283730835864056709609662983759100063333396875182094245046315497525532634764115913236450532733839386139526489824351;

b=1440354387400113353318275132419054375891245413681864837390427511212805748408072838847944629793120889446685643108530381465382074956451566809039119353657601240377236701038904980199109550001860607309184336719930229935342817546146083848277758428344831968440238907935894338978800768226766379;

c=9391500403903773267688655787670246245493629218171544262747638036518222364768797479813561509116827252710188014736501391120827705790025300419608858224262849244058466770043809014864245428958116544162335497194996709759345801074510016208346248254582570123358164225821298549533282498545808644;

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

$\color{#FF4400}{n=6:}$

a=1218343242702905855792264237868803223073090298310121297526752830558323845503910071851999217959704024280699759290559009162035102974023;

b=2250324022012683866886426461942494811141200084921223218461967377588564477616220767789632257358521952443049813799712386367623925971447;

c=20260869859883222379931520298326390700152988332214525711323500132179943287700005601210288797153868533207131302477269470450828233936557.

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

$\color{#FF4400}{n=10:}$

a=221855981602380704196804518854316541759883857932028285581812549404634844243737502744011549757448453135493556098964216532950604590733853450272184987603430882682754171300742698179931849310347;

b=269103113846520710198086599018316928810831097261381335767926880507079911347095440987749703663156874995907158014866846058485318408629957749519665987782327830143454337518378955846463785600977;

c=4862378745380642626737318101484977637219057323564658907686653339599714454790559130946320953938197181210525554039710122136086190642013402927952831079021210585653078786813279351784906397934209.

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

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.

Oleg567
  • 17,740
3

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

$\dfrac{a(c+a)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)}{(a+b)(b+c)(c+a)}$

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 $\gcd(a+b,a+c,b+c)=1$.

Inceptio
  • 7,981
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    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. – celtschk May 26 '13 at 14:24
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    @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.:) – Inceptio May 26 '13 at 14:28
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    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 $2$. – Ivan Loh May 26 '13 at 16:02
  • I considered each pair's $\gcd$ to be $1$[To simplify the case] – Inceptio May 27 '13 at 04:48
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    @IvanLoh Could you please write up your solution? – Potato May 27 '13 at 18:36
  • @Potato Sorry for the delayed response. I was busy at the time and only took a cursory glance at the problem. I did not mean to imply that I had a solution. (Bad phrasing on my part) – Ivan Loh Jun 01 '13 at 13:24