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prove that
with
a positive integerLast edited by tony123 (2008-07-05 06:59:24)
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It's not...
....???
Last edited by Daniel123 (2008-07-05 06:50:39)
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im sory
a positive integerOffline
iirc this problem is considered one of the most difficult problems ever submitted in the IMO...:O
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The equation has only solution N=m=n=1 !
Consider..
(m^2+n^2)/(mn+1)
upon actual division, we get remainder equal to (n/m-m/n) which should be 0 (because its less than mn+1)for N to be an integer!
So, (n/m-m/n)=0
or, m=n
hence, N=2m^2/(m^2+1)
which on division gives the remainder as 2=/=0
therefore, for N to be integer, divisor must divide 2 which is possible only when m^2+1=2 or m=1 (all numbers are positive)!
Which gives N=1 which is of course a square!
any comments are welcome!
BTW, would anyone please be kind enough to take a little pain in teaching me how to write the codes for the above equations to make them look more tidy??
Last edited by ZHero (2008-07-06 00:28:49)
If two or more thoughts intersect, there has to be a point!
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The equation has only solution N=m=n=1 !
Youre wrong.
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Wow!
I always thought that maths was never easy!!
anyways, lemme check if i've got how to write the above..
Last edited by ZHero (2008-07-06 01:39:47)
If two or more thoughts intersect, there has to be a point!
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Heck!
Where am i going wrong??:o
If two or more thoughts intersect, there has to be a point!
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use[ ][/ ] , don t use [/ ] [/ ]
Last edited by tony123 (2008-07-06 01:38:29)
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use[ ][/ ] , don t use [/ ] [/ ]
oooh!
I edited the earlier post and ITS WORKING!
Thanks tony123!
And, i've seen that MOST of your questions are unanswered and they obviously are from Olympiads... Right?
Well, where from do you get these questions?
And if you know the solution to any of those, then do enlighten us too!
The questions are really really amazing and i must say "they add to the beauty of mathematics!!"
If two or more thoughts intersect, there has to be a point!
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Anyways, here's something!
If two or more thoughts intersect, there has to be a point!
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