IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

If

= (n +

(n²- 4))/2 where n is an integer

then prove that

^m = (k+

(k²-4))/2 where k,m are natural numbers.

elastiboysai

hs bhatt,

the gvn eqn implies

is aroot of x^2-nx+1

now 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

hsbhatt

You are only proving that (

^m)^1/m= (n+

(n²-4))/2 =

. We are back to square one.

We should be looking at x^2m- nx^m+1 = 0.

konichiwa2x

Nice one indeed hsbhatt,

Ok first things first. The condition

should be true.

Let

be defined as above. Then,

Fact: for any positive integer

, 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.

Since

, we get:

hsbhatt

The working's fine.

Except that n>=2 need not hold.

You could use induction as follows

Assume

^m + 1/

^m = p (an integer)

Then

^(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.

konichiwa2x

why do u say

need to be true? Wont

become complex otherwise?

yes i am an iit aspirant..12th std student..thanks a lot! i do hope i make it...

hsbhatt

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.

konichiwa2x

oh ok... and thanks for saying that!!! i'll hav to work on chem...

sizzle_through

konichiwa2x nice solun cheers

konichiwa2x

thanks sizzle!

little_genius

konichiwa2x (1302)

konichiwa2x

well, why did u bring up this old post now?

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