1
Rohan Ghosh
·2009-07-14 09:38:28
Doesnt it reduce to the already known equation ??
(a+b)|ab
=>
(a+b)*k=ab
1/a+1/b=1/k {Already known eq.}
further if it is also very good (which is not desired)
n=2a
2a|a^2
=>
2|a
thus n=2*2k=4k
thus numbers divisible by four are excluded in the list :)
39
Dr.House
·2009-07-14 09:44:25
We must now find all numbers of the form 4k which are not very good.
take a careful look.
1
Rohan Ghosh
·2009-07-14 09:58:17
you see any number of the form 4k is very good !!
4k=2k+2k
and
4k|2k*2k
as 2k*2k/4k=k :P
9
Celestine preetham
·2009-07-15 00:12:00
n = a2/λ where a is odd , λ<a
the proof is interesting
juniors this is solvable unlike other very tough olympiad Qs
pls try ....
11
Devil
·2009-07-15 01:45:06
Just a minute, I cud not really get it...
n has to be good...meanng n|ab, so ab=nk.... and n=a+b=2a (Since n is not very good).
So i need a^2/2a=a/2 as an integer which is true for infinitely many such n's?????
9
Celestine preetham
·2009-07-15 01:59:31
yeah i think rohan and i have both taken the opposite !