Rangpur Higher Secondary 2011/5

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Rangpur Higher Secondary 2011/5

Unread post by BdMO » Wed Feb 02, 2011 9:07 pm

Problem 5:
The equation $x^3 +3xy + y^3 = 1$ is solved in nonnegative integers. Find the possible values of $x-y$.

Mehfuj Zahir
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Re: Rangpur Higher Secondary 2011/5

Unread post by Mehfuj Zahir » Thu Feb 03, 2011 12:03 am

x-Y=(1,-1).LOOK OVER THE EQN CAREFULLY AND SEE THE CONDITION

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Moon
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Re: Rangpur Higher Secondary 2011/5

Unread post by Moon » Thu Feb 03, 2011 12:17 am

Please learn to use LaTeX; it will help you to write equations nicely. (It is actually very easy, most of the times you just need to put your equation between two dollar signs.)

Also please don't write with your caps lock on (I mean in uppercase) unless it is absolutely necessary. Writing in uppercase (almost) means you want to shout at somebody. You can use bold or italic text if you need to stress something. Just avoid writing in uppercase.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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bristy1588
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Re: Rangpur Higher Secondary 2011/5

Unread post by bristy1588 » Mon Dec 05, 2011 4:00 pm

Isnt the answer 1 and -1? How come 0 is also the answer?
Bristy Sikder

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*Mahi*
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Re: Rangpur Higher Secondary 2011/5

Unread post by *Mahi* » Tue Dec 06, 2011 12:04 am

$x=-1, y=-1$
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

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bristy1588
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Re: Rangpur Higher Secondary 2011/5

Unread post by bristy1588 » Tue Dec 06, 2011 9:49 pm

Mahi, Question ta dekho okhane bolse non-negative integers, -1 negative integer
Bristy Sikder

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Re: Rangpur Higher Secondary 2011/5

Unread post by nafistiham » Tue Dec 06, 2011 10:30 pm

\[(x,y)=(0,1),(1,0)\]
so,
\[x-y=0,1\]
but, how to prove that?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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