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Rangpur Higher Secondary 2011/5

Posted: Wed Feb 02, 2011 9:07 pm
by BdMO
Problem 5:
The equation $x^3 +3xy + y^3 = 1$ is solved in nonnegative integers. Find the possible values of $x-y$.

Re: Rangpur Higher Secondary 2011/5

Posted: Thu Feb 03, 2011 12:03 am
by Mehfuj Zahir
x-Y=(1,-1).LOOK OVER THE EQN CAREFULLY AND SEE THE CONDITION

Re: Rangpur Higher Secondary 2011/5

Posted: Thu Feb 03, 2011 12:17 am
by Moon
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.

Re: Rangpur Higher Secondary 2011/5

Posted: Mon Dec 05, 2011 4:00 pm
by bristy1588
Isnt the answer 1 and -1? How come 0 is also the answer?

Re: Rangpur Higher Secondary 2011/5

Posted: Tue Dec 06, 2011 12:04 am
by *Mahi*
$x=-1, y=-1$

Re: Rangpur Higher Secondary 2011/5

Posted: Tue Dec 06, 2011 9:49 pm
by bristy1588
Mahi, Question ta dekho okhane bolse non-negative integers, -1 negative integer

Re: Rangpur Higher Secondary 2011/5

Posted: Tue Dec 06, 2011 10:30 pm
by nafistiham
\[(x,y)=(0,1),(1,0)\]
so,
\[x-y=0,1\]
but, how to prove that?