BdMO Online Forum

Show that if a, b, c be integers satisfying \[\frac ab +\frac bc +\frac ca = 3\] then $abc$ is cube of an integer.

Hiks.. hiks ... I made a silly misteke in this problem ..... Ammmmmaaaaa.....
*Mahi* thanks ,

I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.
I did not give a complete solution of this one because of the crises of time.I heard about the problem set at noon in the PMS class.As I had practicals I sat to take the exam at 4 and stopped at 11.That's the reason.But I solved this one later.Everything will be discussed in the next PMS class.And look what I have got today
http://answers.yahoo.com/question/index ... 119AAfBOXz

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

Phlembac Adib Hasan wrote:

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.

In moments like this, sometimes I think, "What should I do? "
Sanzeed, you are wrong, I posted about that yesterday, that you can not WLOG assume $a \geq b \geq c$ here, which is necessary to prove it by rearrangement inequality, because the given equation is cyclic and not symmetric for $a,b,c$.

And Adib, wikipedia says you do not need positive numbers for rearrangement inequality, so I think you are wrong too.

By the way, where did you get the Yahoo answers link? I posted my solutions in general topic on the exam solutions, and I was wondering why the $3$ was not necessary.
(And I also submitted my solutions at 3pm of the first day, if you want to know )

*Mahi* wrote:By the way, where did you get the Yahoo answers link?

,I just googled 'a/b+b/c+c/a perfect cube' and the first result showed me that link.

*Mahi* wrote:

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

Phlembac Adib Hasan wrote:

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post )

Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.

In moments like this, sometimes I think, "What should I do? "
Sanzeed, you are wrong, I posted about that yesterday, that you can not WLOG assume $a \geq b \geq c$ here, which is necessary to prove it by rearrangement inequality, because the given equation is cyclic and not symmetric for $a,b,c$.

And Adib, wikipedia says you do not need positive numbers for rearrangement inequality, so I think you are wrong too.

By the way, where did you get the Yahoo answers link? I posted my solutions in general topic on the exam solutions, and I was wondering why the $3$ was not necessary.
(And I also submitted my solutions at 3pm of the first day, if you want to know )

Re-arrangement inequality-র বিষয়টি clear করার জন্য ধন্যবাদ।আসলেই জানা ছিল না।আর পরশু google search দিয়ে ঐটা পাইসি।আর শোনেন নায়েল ভাইয়াও জানেন যে আমি দুই number-এর complete solution দেই নাই।পরে যদিও পুরা solve করসি, exam-এর মাঝে নিজে যেটুকু পারসি ততটুকুই post করসি।আমার intention মোটেও ঐ রকম না যে কোন website থেকে টুকলিফাই করে post করব।যদি আমি ঐটা exam-এর মাঝেও পাইতাম তবুও কখনও post করতাম না।

Actually I have seen this problem before while I was solving AoPS NT marathon problems last year when I opened a mathlinks account.This one was quite difficult to me and I gave up after some time,then I took a hint(make the expression into a quadratic equation) and finally solved it in an algebric way.(And of course,I informed Nayel vaiya about it).There is a good collection of olympiad level problems there,here's the link:http://www.artofproblemsolving.com/Foru ... start=1140

Mahi vai,I did not understand the thing you wrote for me,would you please make it clear by examples?

I said that, if you assumed $a \geq b \geq c$, then you are wrong. Because the given equation is cyclic, not symmetric. You can find the contradiction by setting $c \geq b \geq a$ or something like that.