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Algebra
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Sairam
Blazing goIITian

Joined: 14 Sep 2007
Posts: 573
17 Jan 2008 21:47:29 IST
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hs bhatt,
the gvn eqn implies
is aroot of x^2-nx+1now to get da eqn whose roots are mth powers of orign eqn, raise x to xpower 1/m
so x^2/m - nx^1/m+1=0 has alpha power m as the roots
and the roots of dis eqn are also of d form k+sqrt(k^2-4)/2
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18 Jan 2008 16:14:56 IST
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Nice one indeed hsbhatt,
Ok first things first. The condition
should be true.
should be true.Let
be defined as above. Then,
be defined as above. Then, Fact: for any positive integer
, the number
is a positive integer.
Proof by induction on
:
, the number
is a positive integer. Proof by induction on
: we have 

Thus
Now on, binomial expansion of the LHS and induction completes the proof.
Let
by above Fact,
is a positive integer.
Let
by above Fact,
is a positive integer. Since
, we get:

, we get: 
18 Jan 2008 18:23:14 IST
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The working's fine.
Except that n>=2 need not hold.
You could use induction as follows
Assume
m + 1/
m = p (an integer)
m + 1/
m = p (an integer)Then
(m+1) + 1/
(m+1) = (
m + 1/
m) (
+1/
) - ((
(m-1) + 1/
(m-1))
(m+1) + 1/
(m+1) = (
m + 1/
m) (
+1/
) - ((
(m-1) + 1/
(m-1))= an integer.
Its a little easier to understand.
By the way, if you are an IIT aspirant, your chances of getting in are pretty good.
18 Jan 2008 21:40:49 IST
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can be complex no harm. Just go through each step of the soln and see if
being complex is any hindrance.U r sure to get IIT man. If you were stock, I would buy you for sure! Just make sure you pay this much attention to Phy and Chem too. They r just as imp.


(n2 - 4))/2 where n is an integer
become complex otherwise?








